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hello-algo/en/docs/chapter_backtracking/subset_sum_problem.md

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---
comments: true
---
# 13.3   Subset sum problem
## 13.3.1   Case without duplicate elements
!!! question
Given an array of positive integers `nums` and a target positive integer `target`, find all possible combinations such that the sum of the elements in the combination equals `target`. The given array has no duplicate elements, and each element can be chosen multiple times. Please return these combinations as a list, which should not contain duplicate combinations.
For example, for the input set $\{3, 4, 5\}$ and target integer $9$, the solutions are $\{3, 3, 3\}, \{4, 5\}$. Note the following two points.
- Elements in the input set can be chosen an unlimited number of times.
- Subsets do not distinguish the order of elements, for example $\{4, 5\}$ and $\{5, 4\}$ are the same subset.
### 1.   Reference permutation solution
Similar to the permutation problem, we can imagine the generation of subsets as a series of choices, updating the "element sum" in real-time during the choice process. When the element sum equals `target`, the subset is recorded in the result list.
Unlike the permutation problem, **elements in this problem can be chosen an unlimited number of times**, thus there is no need to use a `selected` boolean list to record whether an element has been chosen. We can make minor modifications to the permutation code to initially solve the problem:
=== "Python"
```python title="subset_sum_i_naive.py"
def backtrack(
state: list[int],
target: int,
total: int,
choices: list[int],
res: list[list[int]],
):
"""回溯算法：子集和 I"""
# 子集和等于 target 时，记录解
if total == target:
res.append(list(state))
return
# 遍历所有选择
for i in range(len(choices)):
# 剪枝：若子集和超过 target ，则跳过该选择
if total + choices[i] > target:
continue
# 尝试：做出选择，更新元素和 total
state.append(choices[i])
# 进行下一轮选择
backtrack(state, target, total + choices[i], choices, res)
# 回退：撤销选择，恢复到之前的状态
state.pop()
def subset_sum_i_naive(nums: list[int], target: int) -> list[list[int]]:
"""求解子集和 I（包含重复子集）"""
state = [] # 状态（子集）
total = 0 # 子集和
res = [] # 结果列表（子集列表）
backtrack(state, target, total, nums, res)
return res
```
=== "C++"
```cpp title="subset_sum_i_naive.cpp"
/* 回溯算法：子集和 I */
void backtrack(vector<int> &state, int target, int total, vector<int> &choices, vector<vector<int>> &res) {
// 子集和等于 target 时，记录解
if (total == target) {
res.push_back(state);
return;
}
// 遍历所有选择
for (size_t i = 0; i < choices.size(); i++) {
// 剪枝：若子集和超过 target ，则跳过该选择
if (total + choices[i] > target) {
continue;
}
// 尝试：做出选择，更新元素和 total
state.push_back(choices[i]);
// 进行下一轮选择
backtrack(state, target, total + choices[i], choices, res);
// 回退：撤销选择，恢复到之前的状态
state.pop_back();
}
}
/* 求解子集和 I（包含重复子集） */
vector<vector<int>> subsetSumINaive(vector<int> &nums, int target) {
vector<int> state; // 状态（子集）
int total = 0; // 子集和
vector<vector<int>> res; // 结果列表（子集列表）
backtrack(state, target, total, nums, res);
return res;
}
```
=== "Java"
```java title="subset_sum_i_naive.java"
/* 回溯算法：子集和 I */
void backtrack(List<Integer> state, int target, int total, int[] choices, List<List<Integer>> res) {
// 子集和等于 target 时，记录解
if (total == target) {
res.add(new ArrayList<>(state));
return;
}
// 遍历所有选择
for (int i = 0; i < choices.length; i++) {
// 剪枝：若子集和超过 target ，则跳过该选择
if (total + choices[i] > target) {
continue;
}
// 尝试：做出选择，更新元素和 total
state.add(choices[i]);
// 进行下一轮选择
backtrack(state, target, total + choices[i], choices, res);
// 回退：撤销选择，恢复到之前的状态
state.remove(state.size() - 1);
}
}
/* 求解子集和 I（包含重复子集） */
List<List<Integer>> subsetSumINaive(int[] nums, int target) {
List<Integer> state = new ArrayList<>(); // 状态（子集）
int total = 0; // 子集和
List<List<Integer>> res = new ArrayList<>(); // 结果列表（子集列表）
backtrack(state, target, total, nums, res);
return res;
}
```
=== "C#"
```csharp title="subset_sum_i_naive.cs"
/* 回溯算法：子集和 I */
void Backtrack(List<int> state, int target, int total, int[] choices, List<List<int>> res) {
// 子集和等于 target 时，记录解
if (total == target) {
res.Add(new List<int>(state));
return;
}
// 遍历所有选择
for (int i = 0; i < choices.Length; i++) {
// 剪枝：若子集和超过 target ，则跳过该选择
if (total + choices[i] > target) {
continue;
}
// 尝试：做出选择，更新元素和 total
state.Add(choices[i]);
// 进行下一轮选择
Backtrack(state, target, total + choices[i], choices, res);
// 回退：撤销选择，恢复到之前的状态
state.RemoveAt(state.Count - 1);
}
}
/* 求解子集和 I（包含重复子集） */
List<List<int>> SubsetSumINaive(int[] nums, int target) {
List<int> state = []; // 状态（子集）
int total = 0; // 子集和
List<List<int>> res = []; // 结果列表（子集列表）
Backtrack(state, target, total, nums, res);
return res;
}
```
=== "Go"
```go title="subset_sum_i_naive.go"
/* 回溯算法：子集和 I */
func backtrackSubsetSumINaive(total, target int, state, choices *[]int, res *[][]int) {
// 子集和等于 target 时，记录解
if target == total {
newState := append([]int{}, *state...)
*res = append(*res, newState)
return
}
// 遍历所有选择
for i := 0; i < len(*choices); i++ {
// 剪枝：若子集和超过 target ，则跳过该选择
if total+(*choices)[i] > target {
continue
}
// 尝试：做出选择，更新元素和 total
*state = append(*state, (*choices)[i])
// 进行下一轮选择
backtrackSubsetSumINaive(total+(*choices)[i], target, state, choices, res)
// 回退：撤销选择，恢复到之前的状态
*state = (*state)[:len(*state)-1]
}
}
/* 求解子集和 I（包含重复子集） */
func subsetSumINaive(nums []int, target int) [][]int {
state := make([]int, 0) // 状态（子集）
total := 0 // 子集和
res := make([][]int, 0) // 结果列表（子集列表）
backtrackSubsetSumINaive(total, target, &state, &nums, &res)
return res
}
```
=== "Swift"
```swift title="subset_sum_i_naive.swift"
/* 回溯算法：子集和 I */
func backtrack(state: inout [Int], target: Int, total: Int, choices: [Int], res: inout [[Int]]) {
// 子集和等于 target 时，记录解
if total == target {
res.append(state)
return
}
// 遍历所有选择
for i in choices.indices {
// 剪枝：若子集和超过 target ，则跳过该选择
if total + choices[i] > target {
continue
}
// 尝试：做出选择，更新元素和 total
state.append(choices[i])
// 进行下一轮选择
backtrack(state: &state, target: target, total: total + choices[i], choices: choices, res: &res)
// 回退：撤销选择，恢复到之前的状态
state.removeLast()
}
}
/* 求解子集和 I（包含重复子集） */
func subsetSumINaive(nums: [Int], target: Int) -> [[Int]] {
var state: [Int] = [] // 状态（子集）
let total = 0 // 子集和
var res: [[Int]] = [] // 结果列表（子集列表）
backtrack(state: &state, target: target, total: total, choices: nums, res: &res)
return res
}
```
=== "JS"
```javascript title="subset_sum_i_naive.js"
/* 回溯算法：子集和 I */
function backtrack(state, target, total, choices, res) {
// 子集和等于 target 时，记录解
if (total === target) {
res.push([...state]);
return;
}
// 遍历所有选择
for (let i = 0; i < choices.length; i++) {
// 剪枝：若子集和超过 target ，则跳过该选择
if (total + choices[i] > target) {
continue;
}
// 尝试：做出选择，更新元素和 total
state.push(choices[i]);
// 进行下一轮选择
backtrack(state, target, total + choices[i], choices, res);
// 回退：撤销选择，恢复到之前的状态
state.pop();
}
}
/* 求解子集和 I（包含重复子集） */
function subsetSumINaive(nums, target) {
const state = []; // 状态（子集）
const total = 0; // 子集和
const res = []; // 结果列表（子集列表）
backtrack(state, target, total, nums, res);
return res;
}
```
=== "TS"
```typescript title="subset_sum_i_naive.ts"
/* 回溯算法：子集和 I */
function backtrack(
state: number[],
target: number,
total: number,
choices: number[],
res: number[][]
): void {
// 子集和等于 target 时，记录解
if (total === target) {
res.push([...state]);
return;
}
// 遍历所有选择
for (let i = 0; i < choices.length; i++) {
// 剪枝：若子集和超过 target ，则跳过该选择
if (total + choices[i] > target) {
continue;
}
// 尝试：做出选择，更新元素和 total
state.push(choices[i]);
// 进行下一轮选择
backtrack(state, target, total + choices[i], choices, res);
// 回退：撤销选择，恢复到之前的状态
state.pop();
}
}
/* 求解子集和 I（包含重复子集） */
function subsetSumINaive(nums: number[], target: number): number[][] {
const state = []; // 状态（子集）
const total = 0; // 子集和
const res = []; // 结果列表（子集列表）
backtrack(state, target, total, nums, res);
return res;
}
```
=== "Dart"
```dart title="subset_sum_i_naive.dart"
/* 回溯算法：子集和 I */
void backtrack(
List<int> state,
int target,
int total,
List<int> choices,
List<List<int>> res,
) {
// 子集和等于 target 时，记录解
if (total == target) {
res.add(List.from(state));
return;
}
// 遍历所有选择
for (int i = 0; i < choices.length; i++) {
// 剪枝：若子集和超过 target ，则跳过该选择
if (total + choices[i] > target) {
continue;
}
// 尝试：做出选择，更新元素和 total
state.add(choices[i]);
// 进行下一轮选择
backtrack(state, target, total + choices[i], choices, res);
// 回退：撤销选择，恢复到之前的状态
state.removeLast();
}
}
/* 求解子集和 I（包含重复子集） */
List<List<int>> subsetSumINaive(List<int> nums, int target) {
List<int> state = []; // 状态（子集）
int total = 0; // 元素和
List<List<int>> res = []; // 结果列表（子集列表）
backtrack(state, target, total, nums, res);
return res;
}
```
=== "Rust"
```rust title="subset_sum_i_naive.rs"
/* 回溯算法：子集和 I */
fn backtrack(
mut state: Vec<i32>,
target: i32,
total: i32,
choices: &[i32],
res: &mut Vec<Vec<i32>>,
) {
// 子集和等于 target 时，记录解
if total == target {
res.push(state);
return;
}
// 遍历所有选择
for i in 0..choices.len() {
// 剪枝：若子集和超过 target ，则跳过该选择
if total + choices[i] > target {
continue;
}
// 尝试：做出选择，更新元素和 total
state.push(choices[i]);
// 进行下一轮选择
backtrack(state.clone(), target, total + choices[i], choices, res);
// 回退：撤销选择，恢复到之前的状态
state.pop();
}
}
/* 求解子集和 I（包含重复子集） */
fn subset_sum_i_naive(nums: &[i32], target: i32) -> Vec<Vec<i32>> {
let state = Vec::new(); // 状态（子集）
let total = 0; // 子集和
let mut res = Vec::new(); // 结果列表（子集列表）
backtrack(state, target, total, nums, &mut res);
res
}
```
=== "C"
```c title="subset_sum_i_naive.c"
/* 回溯算法：子集和 I */
void backtrack(int target, int total, int *choices, int choicesSize) {
// 子集和等于 target 时，记录解
if (total == target) {
for (int i = 0; i < stateSize; i++) {
res[resSize][i] = state[i];
}
resColSizes[resSize++] = stateSize;
return;
}
// 遍历所有选择
for (int i = 0; i < choicesSize; i++) {
// 剪枝：若子集和超过 target ，则跳过该选择
if (total + choices[i] > target) {
continue;
}
// 尝试：做出选择，更新元素和 total
state[stateSize++] = choices[i];
// 进行下一轮选择
backtrack(target, total + choices[i], choices, choicesSize);
// 回退：撤销选择，恢复到之前的状态
stateSize--;
}
}
/* 求解子集和 I（包含重复子集） */
void subsetSumINaive(int *nums, int numsSize, int target) {
resSize = 0; // 初始化解的数量为0
backtrack(target, 0, nums, numsSize);
}
```
=== "Kotlin"
```kotlin title="subset_sum_i_naive.kt"
/* 回溯算法：子集和 I */
fun backtrack(
state: MutableList<Int>,
target: Int,
total: Int,
choices: IntArray,
res: MutableList<MutableList<Int>?>
) {
// 子集和等于 target 时，记录解
if (total == target) {
res.add(state.toMutableList())
return
}
// 遍历所有选择
for (i in choices.indices) {
// 剪枝：若子集和超过 target ，则跳过该选择
if (total + choices[i] > target) {
continue
}
// 尝试：做出选择，更新元素和 total
state.add(choices[i])
// 进行下一轮选择
backtrack(state, target, total + choices[i], choices, res)
// 回退：撤销选择，恢复到之前的状态
state.removeAt(state.size - 1)
}
}
/* 求解子集和 I（包含重复子集） */
fun subsetSumINaive(nums: IntArray, target: Int): MutableList<MutableList<Int>?> {
val state = mutableListOf<Int>() // 状态（子集）
val total = 0 // 子集和
val res = mutableListOf<MutableList<Int>?>() // 结果列表（子集列表）
backtrack(state, target, total, nums, res)
return res
}
```
=== "Ruby"
```ruby title="subset_sum_i_naive.rb"
[class]{}-[func]{backtrack}
[class]{}-[func]{subset_sum_i_naive}
```
=== "Zig"
```zig title="subset_sum_i_naive.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumINaive}
```
??? pythontutor "Code Visualization"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20backtrack%28%0A%20%20%20%20state%3A%20list%5Bint%5D,%0A%20%20%20%20target%3A%20int,%0A%20%20%20%20total%3A%20int,%0A%20%20%20%20choices%3A%20list%5Bint%5D,%0A%20%20%20%20res%3A%20list%5Blist%5Bint%5D%5D,%0A%29%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%E7%AE%97%E6%B3%95%EF%BC%9A%E5%AD%90%E9%9B%86%E5%92%8C%20I%22%22%22%0A%20%20%20%20%23%20%E5%AD%90%E9%9B%86%E5%92%8C%E7%AD%89%E4%BA%8E%20target%20%E6%97%B6%EF%BC%8C%E8%AE%B0%E5%BD%95%E8%A7%A3%0A%20%20%20%20if%20total%20%3D%3D%20target%3A%0A%20%20%20%20%20%20%20%20res.append%28list%28state%29%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E9%80%89%E6%8B%A9%0A%20%20%20%20for%20i%20in%20range%28len%28choices%29%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%EF%BC%9A%E8%8B%A5%E5%AD%90%E9%9B%86%E5%92%8C%E8%B6%85%E8%BF%87%20target%20%EF%BC%8C%E5%88%99%E8%B7%B3%E8%BF%87%E8%AF%A5%E9%80%89%E6%8B%A9%0A%20%20%20%20%20%20%20%20if%20total%20%2B%20choices%5Bi%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20continue%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%81%9A%E5%87%BA%E9%80%89%E6%8B%A9%EF%BC%8C%E6%9B%B4%E6%96%B0%E5%85%83%E7%B4%A0%E5%92%8C%20total%0A%20%20%20%20%20%20%20%20state.append%28choices%5Bi%5D%29%0A%20%20%20%20%20%20%20%20%23%20%E8%BF%9B%E8%A1%8C%E4%B8%8B%E4%B8%80%E8%BD%AE%E9%80%89%E6%8B%A9%0A%20%20%20%20%20%20%20%20backtrack%28state,%20target,%20total%20%2B%20choices%5Bi%5D,%20choices,%20res%29%0A%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%EF%BC%9A%E6%92%A4%E9%94%80%E9%80%89%E6%8B%A9%EF%BC%8C%E6%81%A2%E5%A4%8D%E5%88%B0%E4%B9%8B%E5%89%8D%E7%9A%84%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20state.pop%28%29%0A%0A%0Adef%20subset_sum_i_naive%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20list%5Blist%5Bint%5D%5D%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E5%AD%90%E9%9B%86%E5%92%8C%20I%EF%BC%88%E5%8C%85%E5%90%AB%E9%87%8D%E5%A4%8D%E5%AD%90%E9%9B%86%EF%BC%89%22%22%22%0A%20%20%20%20state%20%3D%20%5B%5D%20%20%23%20%E7%8A%B6%E6%80%81%EF%BC%88%E5%AD%90%E9%9B%86%EF%BC%89%0A%20%20%20%20total%20%3D%200%20%20%23%20%E5%AD%90%E9%9B%86%E5%92%8C%0A%20%20%20%20res%20%3D%20%5B%5D%20%20%23%20%E7%BB%93%E6%9E%9C%E5%88%97%E8%A1%A8%EF%BC%88%E5%AD%90%E9%9B%86%E5%88%97%E8%A1%A8%EF%BC%89%0A%20%20%20%20backtrack%28state,%20target,%20total,%20nums,%20res%29%0A%20%20%20%20return%20res%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B3,%204,%205%5D%0A%20%20%20%20target%20%3D%209%0A%20%20%20%20res%20%3D%20subset_sum_i_naive%28nums,%20target%29%0A%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%95%B0%E7%BB%84%20nums%20%3D%20%7Bnums%7D,%20target%20%3D%20%7Btarget%7D%22%29%0A%20%20%20%20print%28f%22%E6%89%80%E6%9C%89%E5%92%8C%E7%AD%89%E4%BA%8E%20%7Btarget%7D%20%E7%9A%84%E5%AD%90%E9%9B%86%20res%20%3D%20%7Bres%7D%22%29%0A%20%20%20%20print%28f%22%E8%AF%B7%E6%B3%A8%E6%84%8F%EF%BC%8C%E8%AF%A5%E6%96%B9%E6%B3%95%E8%BE%93%E5%87%BA%E7%9A%84%E7%BB%93%E6%9E%9C%E5%8C%85%E5%90%AB%E9%87%8D%E5%A4%8D%E9%9B%86%E5%90%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=16&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
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Inputting the array $[3, 4, 5]$ and target element $9$ into the above code yields the results $[3, 3, 3], [4, 5], [5, 4]$. **Although it successfully finds all subsets with a sum of $9$, it includes the duplicate subset $[4, 5]$ and $[5, 4]$**.
This is because the search process distinguishes the order of choices, however, subsets do not distinguish the choice order. As shown in the following figure, choosing $4$ before $5$ and choosing $5$ before $4$ are different branches, but correspond to the same subset.
![Subset search and pruning out of bounds](subset_sum_problem.assets/subset_sum_i_naive.png){ class="animation-figure" }
<p align="center"> Figure 13-10 &nbsp; Subset search and pruning out of bounds </p>
To eliminate duplicate subsets, **a straightforward idea is to deduplicate the result list**. However, this method is very inefficient for two reasons.
- When there are many array elements, especially when `target` is large, the search process produces a large number of duplicate subsets.
- Comparing subsets (arrays) for differences is very time-consuming, requiring arrays to be sorted first, then comparing the differences of each element in the arrays.
### 2. &nbsp; Duplicate subset pruning
**We consider deduplication during the search process through pruning**. Observing the following figure, duplicate subsets are generated when choosing array elements in different orders, for example in the following situations.
1. When choosing $3$ in the first round and $4$ in the second round, all subsets containing these two elements are generated, denoted as $[3, 4, \dots]$.
2. Later, when $4$ is chosen in the first round, **the second round should skip $3$** because the subset $[4, 3, \dots]$ generated by this choice completely duplicates the subset from step `1.`.
In the search process, each layer's choices are tried one by one from left to right, so the more to the right a branch is, the more it is pruned.
1. First two rounds choose $3$ and $5$, generating subset $[3, 5, \dots]$.
2. First two rounds choose $4$ and $5$, generating subset $[4, 5, \dots]$.
3. If $5$ is chosen in the first round, **then the second round should skip $3$ and $4$** as the subsets $[5, 3, \dots]$ and $[5, 4, \dots]$ completely duplicate the subsets described in steps `1.` and `2.`.
![Different choice orders leading to duplicate subsets](subset_sum_problem.assets/subset_sum_i_pruning.png){ class="animation-figure" }
<p align="center"> Figure 13-11 &nbsp; Different choice orders leading to duplicate subsets </p>
In summary, given the input array $[x_1, x_2, \dots, x_n]$, the choice sequence in the search process should be $[x_{i_1}, x_{i_2}, \dots, x_{i_m}]$, which needs to satisfy $i_1 \leq i_2 \leq \dots \leq i_m$. **Any choice sequence that does not meet this condition will cause duplicates and should be pruned**.
### 3. &nbsp; Code implementation
To implement this pruning, we initialize the variable `start`, which indicates the starting point for traversal. **After making the choice $x_{i}$, set the next round to start from index $i$**. This will ensure the choice sequence satisfies $i_1 \leq i_2 \leq \dots \leq i_m$, thereby ensuring the uniqueness of the subsets.
Besides, we have made the following two optimizations to the code.
- Before starting the search, sort the array `nums`. In the traversal of all choices, **end the loop directly when the subset sum exceeds `target`** as subsequent elements are larger and their subset sum will definitely exceed `target`.
- Eliminate the element sum variable `total`, **by performing subtraction on `target` to count the element sum**. When `target` equals $0$, record the solution.
=== "Python"
```python title="subset_sum_i.py"
def backtrack(
state: list[int], target: int, choices: list[int], start: int, res: list[list[int]]
):
"""回溯算法：子集和 I"""
# 子集和等于 target 时，记录解
if target == 0:
res.append(list(state))
return
# 遍历所有选择
# 剪枝二：从 start 开始遍历，避免生成重复子集
for i in range(start, len(choices)):
# 剪枝一：若子集和超过 target ，则直接结束循环
# 这是因为数组已排序，后边元素更大，子集和一定超过 target
if target - choices[i] < 0:
break
# 尝试：做出选择，更新 target, start
state.append(choices[i])
# 进行下一轮选择
backtrack(state, target - choices[i], choices, i, res)
# 回退：撤销选择，恢复到之前的状态
state.pop()
def subset_sum_i(nums: list[int], target: int) -> list[list[int]]:
"""求解子集和 I"""
state = [] # 状态（子集）
nums.sort() # 对 nums 进行排序
start = 0 # 遍历起始点
res = [] # 结果列表（子集列表）
backtrack(state, target, nums, start, res)
return res
```
=== "C++"
```cpp title="subset_sum_i.cpp"
/* 回溯算法：子集和 I */
void backtrack(vector<int> &state, int target, vector<int> &choices, int start, vector<vector<int>> &res) {
// 子集和等于 target 时，记录解
if (target == 0) {
res.push_back(state);
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
for (int i = start; i < choices.size(); i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 尝试：做出选择，更新 target, start
state.push_back(choices[i]);
// 进行下一轮选择
backtrack(state, target - choices[i], choices, i, res);
// 回退：撤销选择，恢复到之前的状态
state.pop_back();
}
}
/* 求解子集和 I */
vector<vector<int>> subsetSumI(vector<int> &nums, int target) {
vector<int> state; // 状态（子集）
sort(nums.begin(), nums.end()); // 对 nums 进行排序
int start = 0; // 遍历起始点
vector<vector<int>> res; // 结果列表（子集列表）
backtrack(state, target, nums, start, res);
return res;
}
```
=== "Java"
```java title="subset_sum_i.java"
/* 回溯算法：子集和 I */
void backtrack(List<Integer> state, int target, int[] choices, int start, List<List<Integer>> res) {
// 子集和等于 target 时，记录解
if (target == 0) {
res.add(new ArrayList<>(state));
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
for (int i = start; i < choices.length; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 尝试：做出选择，更新 target, start
state.add(choices[i]);
// 进行下一轮选择
backtrack(state, target - choices[i], choices, i, res);
// 回退：撤销选择，恢复到之前的状态
state.remove(state.size() - 1);
}
}
/* 求解子集和 I */
List<List<Integer>> subsetSumI(int[] nums, int target) {
List<Integer> state = new ArrayList<>(); // 状态（子集）
Arrays.sort(nums); // 对 nums 进行排序
int start = 0; // 遍历起始点
List<List<Integer>> res = new ArrayList<>(); // 结果列表（子集列表）
backtrack(state, target, nums, start, res);
return res;
}
```
=== "C#"
```csharp title="subset_sum_i.cs"
/* 回溯算法：子集和 I */
void Backtrack(List<int> state, int target, int[] choices, int start, List<List<int>> res) {
// 子集和等于 target 时，记录解
if (target == 0) {
res.Add(new List<int>(state));
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
for (int i = start; i < choices.Length; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 尝试：做出选择，更新 target, start
state.Add(choices[i]);
// 进行下一轮选择
Backtrack(state, target - choices[i], choices, i, res);
// 回退：撤销选择，恢复到之前的状态
state.RemoveAt(state.Count - 1);
}
}
/* 求解子集和 I */
List<List<int>> SubsetSumI(int[] nums, int target) {
List<int> state = []; // 状态（子集）
Array.Sort(nums); // 对 nums 进行排序
int start = 0; // 遍历起始点
List<List<int>> res = []; // 结果列表（子集列表）
Backtrack(state, target, nums, start, res);
return res;
}
```
=== "Go"
```go title="subset_sum_i.go"
/* 回溯算法：子集和 I */
func backtrackSubsetSumI(start, target int, state, choices *[]int, res *[][]int) {
// 子集和等于 target 时，记录解
if target == 0 {
newState := append([]int{}, *state...)
*res = append(*res, newState)
return
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
for i := start; i < len(*choices); i++ {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if target-(*choices)[i] < 0 {
break
}
// 尝试：做出选择，更新 target, start
*state = append(*state, (*choices)[i])
// 进行下一轮选择
backtrackSubsetSumI(i, target-(*choices)[i], state, choices, res)
// 回退：撤销选择，恢复到之前的状态
*state = (*state)[:len(*state)-1]
}
}
/* 求解子集和 I */
func subsetSumI(nums []int, target int) [][]int {
state := make([]int, 0) // 状态（子集）
sort.Ints(nums) // 对 nums 进行排序
start := 0 // 遍历起始点
res := make([][]int, 0) // 结果列表（子集列表）
backtrackSubsetSumI(start, target, &state, &nums, &res)
return res
}
```
=== "Swift"
```swift title="subset_sum_i.swift"
/* 回溯算法：子集和 I */
func backtrack(state: inout [Int], target: Int, choices: [Int], start: Int, res: inout [[Int]]) {
// 子集和等于 target 时，记录解
if target == 0 {
res.append(state)
return
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
for i in choices.indices.dropFirst(start) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if target - choices[i] < 0 {
break
}
// 尝试：做出选择，更新 target, start
state.append(choices[i])
// 进行下一轮选择
backtrack(state: &state, target: target - choices[i], choices: choices, start: i, res: &res)
// 回退：撤销选择，恢复到之前的状态
state.removeLast()
}
}
/* 求解子集和 I */
func subsetSumI(nums: [Int], target: Int) -> [[Int]] {
var state: [Int] = [] // 状态（子集）
let nums = nums.sorted() // 对 nums 进行排序
let start = 0 // 遍历起始点
var res: [[Int]] = [] // 结果列表（子集列表）
backtrack(state: &state, target: target, choices: nums, start: start, res: &res)
return res
}
```
=== "JS"
```javascript title="subset_sum_i.js"
/* 回溯算法：子集和 I */
function backtrack(state, target, choices, start, res) {
// 子集和等于 target 时，记录解
if (target === 0) {
res.push([...state]);
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
for (let i = start; i < choices.length; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 尝试：做出选择，更新 target, start
state.push(choices[i]);
// 进行下一轮选择
backtrack(state, target - choices[i], choices, i, res);
// 回退：撤销选择，恢复到之前的状态
state.pop();
}
}
/* 求解子集和 I */
function subsetSumI(nums, target) {
const state = []; // 状态（子集）
nums.sort((a, b) => a - b); // 对 nums 进行排序
const start = 0; // 遍历起始点
const res = []; // 结果列表（子集列表）
backtrack(state, target, nums, start, res);
return res;
}
```
=== "TS"
```typescript title="subset_sum_i.ts"
/* 回溯算法：子集和 I */
function backtrack(
state: number[],
target: number,
choices: number[],
start: number,
res: number[][]
): void {
// 子集和等于 target 时，记录解
if (target === 0) {
res.push([...state]);
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
for (let i = start; i < choices.length; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 尝试：做出选择，更新 target, start
state.push(choices[i]);
// 进行下一轮选择
backtrack(state, target - choices[i], choices, i, res);
// 回退：撤销选择，恢复到之前的状态
state.pop();
}
}
/* 求解子集和 I */
function subsetSumI(nums: number[], target: number): number[][] {
const state = []; // 状态（子集）
nums.sort((a, b) => a - b); // 对 nums 进行排序
const start = 0; // 遍历起始点
const res = []; // 结果列表（子集列表）
backtrack(state, target, nums, start, res);
return res;
}
```
=== "Dart"
```dart title="subset_sum_i.dart"
/* 回溯算法：子集和 I */
void backtrack(
List<int> state,
int target,
List<int> choices,
int start,
List<List<int>> res,
) {
// 子集和等于 target 时，记录解
if (target == 0) {
res.add(List.from(state));
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
for (int i = start; i < choices.length; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 尝试：做出选择，更新 target, start
state.add(choices[i]);
// 进行下一轮选择
backtrack(state, target - choices[i], choices, i, res);
// 回退：撤销选择，恢复到之前的状态
state.removeLast();
}
}
/* 求解子集和 I */
List<List<int>> subsetSumI(List<int> nums, int target) {
List<int> state = []; // 状态（子集）
nums.sort(); // 对 nums 进行排序
int start = 0; // 遍历起始点
List<List<int>> res = []; // 结果列表（子集列表）
backtrack(state, target, nums, start, res);
return res;
}
```
=== "Rust"
```rust title="subset_sum_i.rs"
/* 回溯算法：子集和 I */
fn backtrack(
mut state: Vec<i32>,
target: i32,
choices: &[i32],
start: usize,
res: &mut Vec<Vec<i32>>,
) {
// 子集和等于 target 时，记录解
if target == 0 {
res.push(state);
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
for i in start..choices.len() {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if target - choices[i] < 0 {
break;
}
// 尝试：做出选择，更新 target, start
state.push(choices[i]);
// 进行下一轮选择
backtrack(state.clone(), target - choices[i], choices, i, res);
// 回退：撤销选择，恢复到之前的状态
state.pop();
}
}
/* 求解子集和 I */
fn subset_sum_i(nums: &mut [i32], target: i32) -> Vec<Vec<i32>> {
let state = Vec::new(); // 状态（子集）
nums.sort(); // 对 nums 进行排序
let start = 0; // 遍历起始点
let mut res = Vec::new(); // 结果列表（子集列表）
backtrack(state, target, nums, start, &mut res);
res
}
```
=== "C"
```c title="subset_sum_i.c"
/* 回溯算法：子集和 I */
void backtrack(int target, int *choices, int choicesSize, int start) {
// 子集和等于 target 时，记录解
if (target == 0) {
for (int i = 0; i < stateSize; ++i) {
res[resSize][i] = state[i];
}
resColSizes[resSize++] = stateSize;
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
for (int i = start; i < choicesSize; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 尝试：做出选择，更新 target, start
state[stateSize] = choices[i];
stateSize++;
// 进行下一轮选择
backtrack(target - choices[i], choices, choicesSize, i);
// 回退：撤销选择，恢复到之前的状态
stateSize--;
}
}
/* 求解子集和 I */
void subsetSumI(int *nums, int numsSize, int target) {
qsort(nums, numsSize, sizeof(int), cmp); // 对 nums 进行排序
int start = 0; // 遍历起始点
backtrack(target, nums, numsSize, start);
}
```
=== "Kotlin"
```kotlin title="subset_sum_i.kt"
/* 回溯算法：子集和 I */
fun backtrack(
state: MutableList<Int>,
target: Int,
choices: IntArray,
start: Int,
res: MutableList<MutableList<Int>?>
) {
// 子集和等于 target 时，记录解
if (target == 0) {
res.add(state.toMutableList())
return
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
for (i in start..<choices.size) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break
}
// 尝试：做出选择，更新 target, start
state.add(choices[i])
// 进行下一轮选择
backtrack(state, target - choices[i], choices, i, res)
// 回退：撤销选择，恢复到之前的状态
state.removeAt(state.size - 1)
}
}
/* 求解子集和 I */
fun subsetSumI(nums: IntArray, target: Int): MutableList<MutableList<Int>?> {
val state = mutableListOf<Int>() // 状态（子集）
nums.sort() // 对 nums 进行排序
val start = 0 // 遍历起始点
val res = mutableListOf<MutableList<Int>?>() // 结果列表（子集列表）
backtrack(state, target, nums, start, res)
return res
}
```
=== "Ruby"
```ruby title="subset_sum_i.rb"
[class]{}-[func]{backtrack}
[class]{}-[func]{subset_sum_i}
```
=== "Zig"
```zig title="subset_sum_i.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumI}
```
??? pythontutor "Code Visualization"
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The following figure shows the overall backtracking process after inputting the array $[3, 4, 5]$ and target element $9$ into the above code.
![Subset sum I backtracking process](subset_sum_problem.assets/subset_sum_i.png){ class="animation-figure" }
<p align="center"> Figure 13-12 &nbsp; Subset sum I backtracking process </p>
## 13.3.2 &nbsp; Considering cases with duplicate elements
!!! question
Given an array of positive integers `nums` and a target positive integer `target`, find all possible combinations such that the sum of the elements in the combination equals `target`. **The given array may contain duplicate elements, and each element can only be chosen once**. Please return these combinations as a list, which should not contain duplicate combinations.
Compared to the previous question, **this question's input array may contain duplicate elements**, introducing new problems. For example, given the array $[4, \hat{4}, 5]$ and target element $9$, the existing code's output results in $[4, 5], [\hat{4}, 5]$, resulting in duplicate subsets.
**The reason for this duplication is that equal elements are chosen multiple times in a certain round**. In the following figure, the first round has three choices, two of which are $4$, generating two duplicate search branches, thus outputting duplicate subsets; similarly, the two $4$s in the second round also produce duplicate subsets.
![Duplicate subsets caused by equal elements](subset_sum_problem.assets/subset_sum_ii_repeat.png){ class="animation-figure" }
<p align="center"> Figure 13-13 &nbsp; Duplicate subsets caused by equal elements </p>
### 1. &nbsp; Equal element pruning
To solve this issue, **we need to limit equal elements to being chosen only once per round**. The implementation is quite clever: since the array is sorted, equal elements are adjacent. This means that in a certain round of choices, if the current element is equal to its left-hand element, it means it has already been chosen, so skip the current element directly.
At the same time, **this question stipulates that each array element can only be chosen once**. Fortunately, we can also use the variable `start` to meet this constraint: after making the choice $x_{i}$, set the next round to start from index $i + 1$ going forward. This not only eliminates duplicate subsets but also avoids repeated selection of elements.
### 2. &nbsp; Code implementation
=== "Python"
```python title="subset_sum_ii.py"
def backtrack(
state: list[int], target: int, choices: list[int], start: int, res: list[list[int]]
):
"""回溯算法：子集和 II"""
# 子集和等于 target 时，记录解
if target == 0:
res.append(list(state))
return
# 遍历所有选择
# 剪枝二：从 start 开始遍历，避免生成重复子集
# 剪枝三：从 start 开始遍历，避免重复选择同一元素
for i in range(start, len(choices)):
# 剪枝一：若子集和超过 target ，则直接结束循环
# 这是因为数组已排序，后边元素更大，子集和一定超过 target
if target - choices[i] < 0:
break
# 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if i > start and choices[i] == choices[i - 1]:
continue
# 尝试：做出选择，更新 target, start
state.append(choices[i])
# 进行下一轮选择
backtrack(state, target - choices[i], choices, i + 1, res)
# 回退：撤销选择，恢复到之前的状态
state.pop()
def subset_sum_ii(nums: list[int], target: int) -> list[list[int]]:
"""求解子集和 II"""
state = [] # 状态（子集）
nums.sort() # 对 nums 进行排序
start = 0 # 遍历起始点
res = [] # 结果列表（子集列表）
backtrack(state, target, nums, start, res)
return res
```
=== "C++"
```cpp title="subset_sum_ii.cpp"
/* 回溯算法：子集和 II */
void backtrack(vector<int> &state, int target, vector<int> &choices, int start, vector<vector<int>> &res) {
// 子集和等于 target 时，记录解
if (target == 0) {
res.push_back(state);
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
// 剪枝三：从 start 开始遍历，避免重复选择同一元素
for (int i = start; i < choices.size(); i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if (i > start && choices[i] == choices[i - 1]) {
continue;
}
// 尝试：做出选择，更新 target, start
state.push_back(choices[i]);
// 进行下一轮选择
backtrack(state, target - choices[i], choices, i + 1, res);
// 回退：撤销选择，恢复到之前的状态
state.pop_back();
}
}
/* 求解子集和 II */
vector<vector<int>> subsetSumII(vector<int> &nums, int target) {
vector<int> state; // 状态（子集）
sort(nums.begin(), nums.end()); // 对 nums 进行排序
int start = 0; // 遍历起始点
vector<vector<int>> res; // 结果列表（子集列表）
backtrack(state, target, nums, start, res);
return res;
}
```
=== "Java"
```java title="subset_sum_ii.java"
/* 回溯算法：子集和 II */
void backtrack(List<Integer> state, int target, int[] choices, int start, List<List<Integer>> res) {
// 子集和等于 target 时，记录解
if (target == 0) {
res.add(new ArrayList<>(state));
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
// 剪枝三：从 start 开始遍历，避免重复选择同一元素
for (int i = start; i < choices.length; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if (i > start && choices[i] == choices[i - 1]) {
continue;
}
// 尝试：做出选择，更新 target, start
state.add(choices[i]);
// 进行下一轮选择
backtrack(state, target - choices[i], choices, i + 1, res);
// 回退：撤销选择，恢复到之前的状态
state.remove(state.size() - 1);
}
}
/* 求解子集和 II */
List<List<Integer>> subsetSumII(int[] nums, int target) {
List<Integer> state = new ArrayList<>(); // 状态（子集）
Arrays.sort(nums); // 对 nums 进行排序
int start = 0; // 遍历起始点
List<List<Integer>> res = new ArrayList<>(); // 结果列表（子集列表）
backtrack(state, target, nums, start, res);
return res;
}
```
=== "C#"
```csharp title="subset_sum_ii.cs"
/* 回溯算法：子集和 II */
void Backtrack(List<int> state, int target, int[] choices, int start, List<List<int>> res) {
// 子集和等于 target 时，记录解
if (target == 0) {
res.Add(new List<int>(state));
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
// 剪枝三：从 start 开始遍历，避免重复选择同一元素
for (int i = start; i < choices.Length; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if (i > start && choices[i] == choices[i - 1]) {
continue;
}
// 尝试：做出选择，更新 target, start
state.Add(choices[i]);
// 进行下一轮选择
Backtrack(state, target - choices[i], choices, i + 1, res);
// 回退：撤销选择，恢复到之前的状态
state.RemoveAt(state.Count - 1);
}
}
/* 求解子集和 II */
List<List<int>> SubsetSumII(int[] nums, int target) {
List<int> state = []; // 状态（子集）
Array.Sort(nums); // 对 nums 进行排序
int start = 0; // 遍历起始点
List<List<int>> res = []; // 结果列表（子集列表）
Backtrack(state, target, nums, start, res);
return res;
}
```
=== "Go"
```go title="subset_sum_ii.go"
/* 回溯算法：子集和 II */
func backtrackSubsetSumII(start, target int, state, choices *[]int, res *[][]int) {
// 子集和等于 target 时，记录解
if target == 0 {
newState := append([]int{}, *state...)
*res = append(*res, newState)
return
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
// 剪枝三：从 start 开始遍历，避免重复选择同一元素
for i := start; i < len(*choices); i++ {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if target-(*choices)[i] < 0 {
break
}
// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if i > start && (*choices)[i] == (*choices)[i-1] {
continue
}
// 尝试：做出选择，更新 target, start
*state = append(*state, (*choices)[i])
// 进行下一轮选择
backtrackSubsetSumII(i+1, target-(*choices)[i], state, choices, res)
// 回退：撤销选择，恢复到之前的状态
*state = (*state)[:len(*state)-1]
}
}
/* 求解子集和 II */
func subsetSumII(nums []int, target int) [][]int {
state := make([]int, 0) // 状态（子集）
sort.Ints(nums) // 对 nums 进行排序
start := 0 // 遍历起始点
res := make([][]int, 0) // 结果列表（子集列表）
backtrackSubsetSumII(start, target, &state, &nums, &res)
return res
}
```
=== "Swift"
```swift title="subset_sum_ii.swift"
/* 回溯算法：子集和 II */
func backtrack(state: inout [Int], target: Int, choices: [Int], start: Int, res: inout [[Int]]) {
// 子集和等于 target 时，记录解
if target == 0 {
res.append(state)
return
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
// 剪枝三：从 start 开始遍历，避免重复选择同一元素
for i in choices.indices.dropFirst(start) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if target - choices[i] < 0 {
break
}
// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if i > start, choices[i] == choices[i - 1] {
continue
}
// 尝试：做出选择，更新 target, start
state.append(choices[i])
// 进行下一轮选择
backtrack(state: &state, target: target - choices[i], choices: choices, start: i + 1, res: &res)
// 回退：撤销选择，恢复到之前的状态
state.removeLast()
}
}
/* 求解子集和 II */
func subsetSumII(nums: [Int], target: Int) -> [[Int]] {
var state: [Int] = [] // 状态（子集）
let nums = nums.sorted() // 对 nums 进行排序
let start = 0 // 遍历起始点
var res: [[Int]] = [] // 结果列表（子集列表）
backtrack(state: &state, target: target, choices: nums, start: start, res: &res)
return res
}
```
=== "JS"
```javascript title="subset_sum_ii.js"
/* 回溯算法：子集和 II */
function backtrack(state, target, choices, start, res) {
// 子集和等于 target 时，记录解
if (target === 0) {
res.push([...state]);
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
// 剪枝三：从 start 开始遍历，避免重复选择同一元素
for (let i = start; i < choices.length; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if (i > start && choices[i] === choices[i - 1]) {
continue;
}
// 尝试：做出选择，更新 target, start
state.push(choices[i]);
// 进行下一轮选择
backtrack(state, target - choices[i], choices, i + 1, res);
// 回退：撤销选择，恢复到之前的状态
state.pop();
}
}
/* 求解子集和 II */
function subsetSumII(nums, target) {
const state = []; // 状态（子集）
nums.sort((a, b) => a - b); // 对 nums 进行排序
const start = 0; // 遍历起始点
const res = []; // 结果列表（子集列表）
backtrack(state, target, nums, start, res);
return res;
}
```
=== "TS"
```typescript title="subset_sum_ii.ts"
/* 回溯算法：子集和 II */
function backtrack(
state: number[],
target: number,
choices: number[],
start: number,
res: number[][]
): void {
// 子集和等于 target 时，记录解
if (target === 0) {
res.push([...state]);
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
// 剪枝三：从 start 开始遍历，避免重复选择同一元素
for (let i = start; i < choices.length; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if (i > start && choices[i] === choices[i - 1]) {
continue;
}
// 尝试：做出选择，更新 target, start
state.push(choices[i]);
// 进行下一轮选择
backtrack(state, target - choices[i], choices, i + 1, res);
// 回退：撤销选择，恢复到之前的状态
state.pop();
}
}
/* 求解子集和 II */
function subsetSumII(nums: number[], target: number): number[][] {
const state = []; // 状态（子集）
nums.sort((a, b) => a - b); // 对 nums 进行排序
const start = 0; // 遍历起始点
const res = []; // 结果列表（子集列表）
backtrack(state, target, nums, start, res);
return res;
}
```
=== "Dart"
```dart title="subset_sum_ii.dart"
/* 回溯算法：子集和 II */
void backtrack(
List<int> state,
int target,
List<int> choices,
int start,
List<List<int>> res,
) {
// 子集和等于 target 时，记录解
if (target == 0) {
res.add(List.from(state));
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
// 剪枝三：从 start 开始遍历，避免重复选择同一元素
for (int i = start; i < choices.length; i++) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break;
}
// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if (i > start && choices[i] == choices[i - 1]) {
continue;
}
// 尝试：做出选择，更新 target, start
state.add(choices[i]);
// 进行下一轮选择
backtrack(state, target - choices[i], choices, i + 1, res);
// 回退：撤销选择，恢复到之前的状态
state.removeLast();
}
}
/* 求解子集和 II */
List<List<int>> subsetSumII(List<int> nums, int target) {
List<int> state = []; // 状态（子集）
nums.sort(); // 对 nums 进行排序
int start = 0; // 遍历起始点
List<List<int>> res = []; // 结果列表（子集列表）
backtrack(state, target, nums, start, res);
return res;
}
```
=== "Rust"
```rust title="subset_sum_ii.rs"
/* 回溯算法：子集和 II */
fn backtrack(
mut state: Vec<i32>,
target: i32,
choices: &[i32],
start: usize,
res: &mut Vec<Vec<i32>>,
) {
// 子集和等于 target 时，记录解
if target == 0 {
res.push(state);
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
// 剪枝三：从 start 开始遍历，避免重复选择同一元素
for i in start..choices.len() {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if target - choices[i] < 0 {
break;
}
// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if i > start && choices[i] == choices[i - 1] {
continue;
}
// 尝试：做出选择，更新 target, start
state.push(choices[i]);
// 进行下一轮选择
backtrack(state.clone(), target - choices[i], choices, i, res);
// 回退：撤销选择，恢复到之前的状态
state.pop();
}
}
/* 求解子集和 II */
fn subset_sum_ii(nums: &mut [i32], target: i32) -> Vec<Vec<i32>> {
let state = Vec::new(); // 状态（子集）
nums.sort(); // 对 nums 进行排序
let start = 0; // 遍历起始点
let mut res = Vec::new(); // 结果列表（子集列表）
backtrack(state, target, nums, start, &mut res);
res
}
```
=== "C"
```c title="subset_sum_ii.c"
/* 回溯算法：子集和 II */
void backtrack(int target, int *choices, int choicesSize, int start) {
// 子集和等于 target 时，记录解
if (target == 0) {
for (int i = 0; i < stateSize; i++) {
res[resSize][i] = state[i];
}
resColSizes[resSize++] = stateSize;
return;
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
// 剪枝三：从 start 开始遍历，避免重复选择同一元素
for (int i = start; i < choicesSize; i++) {
// 剪枝一：若子集和超过 target ，则直接跳过
if (target - choices[i] < 0) {
continue;
}
// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if (i > start && choices[i] == choices[i - 1]) {
continue;
}
// 尝试：做出选择，更新 target, start
state[stateSize] = choices[i];
stateSize++;
// 进行下一轮选择
backtrack(target - choices[i], choices, choicesSize, i + 1);
// 回退：撤销选择，恢复到之前的状态
stateSize--;
}
}
/* 求解子集和 II */
void subsetSumII(int *nums, int numsSize, int target) {
// 对 nums 进行排序
qsort(nums, numsSize, sizeof(int), cmp);
// 开始回溯
backtrack(target, nums, numsSize, 0);
}
```
=== "Kotlin"
```kotlin title="subset_sum_ii.kt"
/* 回溯算法：子集和 II */
fun backtrack(
state: MutableList<Int>,
target: Int,
choices: IntArray,
start: Int,
res: MutableList<MutableList<Int>?>
) {
// 子集和等于 target 时，记录解
if (target == 0) {
res.add(state.toMutableList())
return
}
// 遍历所有选择
// 剪枝二：从 start 开始遍历，避免生成重复子集
// 剪枝三：从 start 开始遍历，避免重复选择同一元素
for (i in start..<choices.size) {
// 剪枝一：若子集和超过 target ，则直接结束循环
// 这是因为数组已排序，后边元素更大，子集和一定超过 target
if (target - choices[i] < 0) {
break
}
// 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过
if (i > start && choices[i] == choices[i - 1]) {
continue
}
// 尝试：做出选择，更新 target, start
state.add(choices[i])
// 进行下一轮选择
backtrack(state, target - choices[i], choices, i + 1, res)
// 回退：撤销选择，恢复到之前的状态
state.removeAt(state.size - 1)
}
}
/* 求解子集和 II */
fun subsetSumII(nums: IntArray, target: Int): MutableList<MutableList<Int>?> {
val state = mutableListOf<Int>() // 状态（子集）
nums.sort() // 对 nums 进行排序
val start = 0 // 遍历起始点
val res = mutableListOf<MutableList<Int>?>() // 结果列表（子集列表）
backtrack(state, target, nums, start, res)
return res
}
```
=== "Ruby"
```ruby title="subset_sum_ii.rb"
[class]{}-[func]{backtrack}
[class]{}-[func]{subset_sum_ii}
```
=== "Zig"
```zig title="subset_sum_ii.zig"
[class]{}-[func]{backtrack}
[class]{}-[func]{subsetSumII}
```
??? pythontutor "Code Visualization"
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<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20backtrack%28%0A%20%20%20%20state%3A%20list%5Bint%5D,%20target%3A%20int,%20choices%3A%20list%5Bint%5D,%20start%3A%20int,%20res%3A%20list%5Blist%5Bint%5D%5D%0A%29%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%E7%AE%97%E6%B3%95%EF%BC%9A%E5%AD%90%E9%9B%86%E5%92%8C%20II%22%22%22%0A%20%20%20%20%23%20%E5%AD%90%E9%9B%86%E5%92%8C%E7%AD%89%E4%BA%8E%20target%20%E6%97%B6%EF%BC%8C%E8%AE%B0%E5%BD%95%E8%A7%A3%0A%20%20%20%20if%20target%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20res.append%28list%28state%29%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E9%80%89%E6%8B%A9%0A%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E4%BA%8C%EF%BC%9A%E4%BB%8E%20start%20%E5%BC%80%E5%A7%8B%E9%81%8D%E5%8E%86%EF%BC%8C%E9%81%BF%E5%85%8D%E7%94%9F%E6%88%90%E9%87%8D%E5%A4%8D%E5%AD%90%E9%9B%86%0A%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E4%B8%89%EF%BC%9A%E4%BB%8E%20start%20%E5%BC%80%E5%A7%8B%E9%81%8D%E5%8E%86%EF%BC%8C%E9%81%BF%E5%85%8D%E9%87%8D%E5%A4%8D%E9%80%89%E6%8B%A9%E5%90%8C%E4%B8%80%E5%85%83%E7%B4%A0%0A%20%20%20%20for%20i%20in%20range%28start,%20len%28choices%29%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E4%B8%80%EF%BC%9A%E8%8B%A5%E5%AD%90%E9%9B%86%E5%92%8C%E8%B6%85%E8%BF%87%20target%20%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E7%BB%93%E6%9D%9F%E5%BE%AA%E7%8E%AF%0A%20%20%20%20%20%20%20%20%23%20%E8%BF%99%E6%98%AF%E5%9B%A0%E4%B8%BA%E6%95%B0%E7%BB%84%E5%B7%B2%E6%8E%92%E5%BA%8F%EF%BC%8C%E5%90%8E%E8%BE%B9%E5%85%83%E7%B4%A0%E6%9B%B4%E5%A4%A7%EF%BC%8C%E5%AD%90%E9%9B%86%E5%92%8C%E4%B8%80%E5%AE%9A%E8%B6%85%E8%BF%87%20target%0A%20%20%20%20%20%20%20%20if%20target%20-%20choices%5Bi%5D%20%3C%200%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20break%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E5%9B%9B%EF%BC%9A%E5%A6%82%E6%9E%9C%E8%AF%A5%E5%85%83%E7%B4%A0%E4%B8%8E%E5%B7%A6%E8%BE%B9%E5%85%83%E7%B4%A0%E7%9B%B8%E7%AD%89%EF%BC%8C%E8%AF%B4%E6%98%8E%E8%AF%A5%E6%90%9C%E7%B4%A2%E5%88%86%E6%94%AF%E9%87%8D%E5%A4%8D%EF%BC%8C%E7%9B%B4%E6%8E%A5%E8%B7%B3%E8%BF%87%0A%20%20%20%20%20%20%20%20if%20i%20%3E%20start%20and%20choices%5Bi%5D%20%3D%3D%20choices%5Bi%20-%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20continue%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%81%9A%E5%87%BA%E9%80%89%E6%8B%A9%EF%BC%8C%E6%9B%B4%E6%96%B0%20target,%20start%0A%20%20%20%20%20%20%20%20state.append%28choices%5Bi%5D%29%0A%20%20%20%20%20%20%20%20%23%20%E8%BF%9B%E8%A1%8C%E4%B8%8B%E4%B8%80%E8%BD%AE%E9%80%89%E6%8B%A9%0A%20%20%20%20%20%20%20%20backtrack%28state,%20target%20-%20choices%5Bi%5D,%20choices,%20i%20%2B%201,%20res%29%0A%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%EF%BC%9A%E6%92%A4%E9%94%80%E9%80%89%E6%8B%A9%EF%BC%8C%E6%81%A2%E5%A4%8D%E5%88%B0%E4%B9%8B%E5%89%8D%E7%9A%84%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20state.pop%28%29%0A%0A%0Adef%20subset_sum_ii%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20list%5Blist%5Bint%5D%5D%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E5%AD%90%E9%9B%86%E5%92%8C%20II%22%22%22%0A%20%20%20%20state%20%3D%20%5B%5D%20%20%23%20%E7%8A%B6%E6%80%81%EF%BC%88%E5%AD%90%E9%9B%86%EF%BC%89%0A%20%20%20%20nums.sort%28%29%20%20%23%20%E5%AF%B9%20nums%20%E8%BF%9B%E8%A1%8C%E6%8E%92%E5%BA%8F%0A%20%20%20%20start%20%3D%200%20%20%23%20%E9%81%8D%E5%8E%86%E8%B5%B7%E5%A7%8B%E7%82%B9%0A%20%20%20%20res%20%3D%20%5B%5D%20%20%23%20%E7%BB%93%E6%9E%9C%E5%88%97%E8%A1%A8%EF%BC%88%E5%AD%90%E9%9B%86%E5%88%97%E8%A1%A8%EF%BC%89%0A%20%20%20%20backtrack%28state,%20target,%20nums,%20start,%20res%29%0A%20%20%20%20return%20res%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B4,%204,%205%5D%0A%20%20%20%20target%20%3D%209%0A%20%20%20%20res%20%3D%20subset_sum_ii%28nums,%20target%29%0A%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%95%B0%E7%BB%84%20nums%20%3D%20%7Bnums%7D,%20target%20%3D%20%7Btarget%7D%22%29%0A%20%20%20%20print%28f%22%E6%89%80%E6%9C%89%E5%92%8C%E7%AD%89%E4%BA%8E%20%7Btarget%7D%20%E7%9A%84%E5%AD%90%E9%9B%86%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cum
The following figure shows the backtracking process for the array $[4, 4, 5]$ and target element $9$, including four types of pruning operations. Please combine the illustration with the code comments to understand the entire search process and how each type of pruning operation works.
![Subset sum II backtracking process](subset_sum_problem.assets/subset_sum_ii.png){ class="animation-figure" }
<p align="center"> Figure 13-14 &nbsp; Subset sum II backtracking process </p>