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10.4   雜湊最佳化策略

在演算法題中，我們常透過將線性查詢替換為雜湊查詢來降低演算法的時間複雜度。我們藉助一個演算法題來加深理解。

!!! question

給定一個整數陣列 `nums` 和一個目標元素 `target` ，請在陣列中搜索“和”為 `target` 的兩個元素，並返回它們的陣列索引。返回任意一個解即可。

10.4.1   線性查詢：以時間換空間

考慮直接走訪所有可能的組合。如圖 10-9 所示，我們開啟一個兩層迴圈，在每輪中判斷兩個整數的和是否為 target ，若是，則返回它們的索引。

線性查詢求解兩數之和{ class="animation-figure" }

圖 10-9   線性查詢求解兩數之和

程式碼如下所示：

=== "Python"

```python title="two_sum.py"
def two_sum_brute_force(nums: list[int], target: int) -> list[int]:
    """方法一：暴力列舉"""
    # 兩層迴圈，時間複雜度為 O(n^2)
    for i in range(len(nums) - 1):
        for j in range(i + 1, len(nums)):
            if nums[i] + nums[j] == target:
                return [i, j]
    return []
```

=== "C++"

```cpp title="two_sum.cpp"
/* 方法一：暴力列舉 */
vector<int> twoSumBruteForce(vector<int> &nums, int target) {
    int size = nums.size();
    // 兩層迴圈，時間複雜度為 O(n^2)
    for (int i = 0; i < size - 1; i++) {
        for (int j = i + 1; j < size; j++) {
            if (nums[i] + nums[j] == target)
                return {i, j};
        }
    }
    return {};
}
```

=== "Java"

```java title="two_sum.java"
/* 方法一：暴力列舉 */
int[] twoSumBruteForce(int[] nums, int target) {
    int size = nums.length;
    // 兩層迴圈，時間複雜度為 O(n^2)
    for (int i = 0; i < size - 1; i++) {
        for (int j = i + 1; j < size; j++) {
            if (nums[i] + nums[j] == target)
                return new int[] { i, j };
        }
    }
    return new int[0];
}
```

=== "C#"

```csharp title="two_sum.cs"
/* 方法一：暴力列舉 */
int[] TwoSumBruteForce(int[] nums, int target) {
    int size = nums.Length;
    // 兩層迴圈，時間複雜度為 O(n^2)
    for (int i = 0; i < size - 1; i++) {
        for (int j = i + 1; j < size; j++) {
            if (nums[i] + nums[j] == target)
                return [i, j];
        }
    }
    return [];
}
```

=== "Go"

```go title="two_sum.go"
/* 方法一：暴力列舉 */
func twoSumBruteForce(nums []int, target int) []int {
    size := len(nums)
    // 兩層迴圈，時間複雜度為 O(n^2)
    for i := 0; i < size-1; i++ {
        for j := i + 1; j < size; j++ {
            if nums[i]+nums[j] == target {
                return []int{i, j}
            }
        }
    }
    return nil
}
```

=== "Swift"

```swift title="two_sum.swift"
/* 方法一：暴力列舉 */
func twoSumBruteForce(nums: [Int], target: Int) -> [Int] {
    // 兩層迴圈，時間複雜度為 O(n^2)
    for i in nums.indices.dropLast() {
        for j in nums.indices.dropFirst(i + 1) {
            if nums[i] + nums[j] == target {
                return [i, j]
            }
        }
    }
    return [0]
}
```

=== "JS"

```javascript title="two_sum.js"
/* 方法一：暴力列舉 */
function twoSumBruteForce(nums, target) {
    const n = nums.length;
    // 兩層迴圈，時間複雜度為 O(n^2)
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            if (nums[i] + nums[j] === target) {
                return [i, j];
            }
        }
    }
    return [];
}
```

=== "TS"

```typescript title="two_sum.ts"
/* 方法一：暴力列舉 */
function twoSumBruteForce(nums: number[], target: number): number[] {
    const n = nums.length;
    // 兩層迴圈，時間複雜度為 O(n^2)
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            if (nums[i] + nums[j] === target) {
                return [i, j];
            }
        }
    }
    return [];
}
```

=== "Dart"

```dart title="two_sum.dart"
/* 方法一： 暴力列舉 */
List<int> twoSumBruteForce(List<int> nums, int target) {
  int size = nums.length;
  // 兩層迴圈，時間複雜度為 O(n^2)
  for (var i = 0; i < size - 1; i++) {
    for (var j = i + 1; j < size; j++) {
      if (nums[i] + nums[j] == target) return [i, j];
    }
  }
  return [0];
}
```

=== "Rust"

```rust title="two_sum.rs"
/* 方法一：暴力列舉 */
pub fn two_sum_brute_force(nums: &Vec<i32>, target: i32) -> Option<Vec<i32>> {
    let size = nums.len();
    // 兩層迴圈，時間複雜度為 O(n^2)
    for i in 0..size - 1 {
        for j in i + 1..size {
            if nums[i] + nums[j] == target {
                return Some(vec![i as i32, j as i32]);
            }
        }
    }
    None
}
```

=== "C"

```c title="two_sum.c"
/* 方法一：暴力列舉 */
int *twoSumBruteForce(int *nums, int numsSize, int target, int *returnSize) {
    for (int i = 0; i < numsSize; ++i) {
        for (int j = i + 1; j < numsSize; ++j) {
            if (nums[i] + nums[j] == target) {
                int *res = malloc(sizeof(int) * 2);
                res[0] = i, res[1] = j;
                *returnSize = 2;
                return res;
            }
        }
    }
    *returnSize = 0;
    return NULL;
}
```

=== "Kotlin"

```kotlin title="two_sum.kt"
/* 方法一：暴力列舉 */
fun twoSumBruteForce(nums: IntArray, target: Int): IntArray {
    val size = nums.size
    // 兩層迴圈，時間複雜度為 O(n^2)
    for (i in 0..<size - 1) {
        for (j in i + 1..<size) {
            if (nums[i] + nums[j] == target) return intArrayOf(i, j)
        }
    }
    return IntArray(0)
}
```

=== "Ruby"

```ruby title="two_sum.rb"
### 方法一：暴力列舉 ###
def two_sum_brute_force(nums, target)
  # 兩層迴圈，時間複雜度為 O(n^2)
  for i in 0...(nums.length - 1)
    for j in (i + 1)...nums.length
      return [i, j] if nums[i] + nums[j] == target
    end
  end

  []
end
```

=== "Zig"

```zig title="two_sum.zig"
// 方法一：暴力列舉
fn twoSumBruteForce(nums: []i32, target: i32) ?[2]i32 {
    var size: usize = nums.len;
    var i: usize = 0;
    // 兩層迴圈，時間複雜度為 O(n^2)
    while (i < size - 1) : (i += 1) {
        var j = i + 1;
        while (j < size) : (j += 1) {
            if (nums[i] + nums[j] == target) {
                return [_]i32{@intCast(i), @intCast(j)};
            }
        }
    }
    return null;
}
```

??? pythontutor "視覺化執行"

<div style="height: 441px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20two_sum_brute_force%28nums%3A%20list%5Bint%5D%2C%20target%3A%20int%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%22%22%22%E6%96%B9%E6%B3%95%E4%B8%80%EF%BC%9A%E6%9A%B4%E5%8A%9B%E5%88%97%E8%88%89%22%22%22%0A%20%20%20%20%23%20%E5%85%A9%E5%B1%A4%E8%BF%B4%E5%9C%88%EF%BC%8C%E6%99%82%E9%96%93%E8%A4%87%E9%9B%9C%E5%BA%A6%E7%82%BA%20O%28n%5E2%29%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%20-%201%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%28i%20%2B%201%2C%20len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20nums%5Bi%5D%20%2B%20nums%5Bj%5D%20%3D%3D%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20return%20%5Bi%2C%20j%5D%0A%20%20%20%20return%20%5B%5D%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%5B2%2C%207%2C%2011%2C%2015%5D%0A%20%20%20%20target%20%3D%2013%0A%20%20%20%20res%20%3D%20two_sum_brute_force%28nums%2C%20target%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20two_sum_brute_force%28nums%3A%20list%5Bint%5D%2C%20target%3A%20int%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%22%22%22%E6%96%B9%E6%B3%95%E4%B8%80%EF%BC%9A%E6%9A%B4%E5%8A%9B%E5%88%97%E8%88%89%22%22%22%0A%20%20%20%20%23%20%E5%85%A9%E5%B1%A4%E8%BF%B4%E5%9C%88%EF%BC%8C%E6%99%82%E9%96%93%E8%A4%87%E9%9B%9C%E5%BA%A6%E7%82%BA%20O%28n%5E2%29%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%20-%201%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%28i%20%2B%201%2C%20len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20nums%5Bi%5D%20%2B%20nums%5Bj%5D%20%3D%3D%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20return%20%5Bi%2C%20j%5D%0A%20%20%20%20return%20%5B%5D%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%5B2%2C%207%2C%2011%2C%2015%5D%0A%20%20%20%20target%20%3D%2013%0A%20%20%20%20res%20%3D%20two_sum_brute_force%28nums%2C%20target%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>

此方法的時間複雜度為 O(n^2) ，空間複雜度為 O(1) ，在大資料量下非常耗時。

10.4.2   雜湊查詢：以空間換時間

考慮藉助一個雜湊表，鍵值對分別為陣列元素和元素索引。迴圈走訪陣列，每輪執行圖 10-10 所示的步驟。

  1. 判斷數字 target - nums[i] 是否在雜湊表中，若是，則直接返回這兩個元素的索引。
  2. 將鍵值對 nums[i] 和索引 i 新增進雜湊表。

=== "<1>" 輔助雜湊表求解兩數之和{ class="animation-figure" }

=== "<2>" two_sum_hashtable_step2{ class="animation-figure" }

=== "<3>" two_sum_hashtable_step3{ class="animation-figure" }

圖 10-10   輔助雜湊表求解兩數之和

實現程式碼如下所示，僅需單層迴圈即可：

=== "Python"

```python title="two_sum.py"
def two_sum_hash_table(nums: list[int], target: int) -> list[int]:
    """方法二：輔助雜湊表"""
    # 輔助雜湊表，空間複雜度為 O(n)
    dic = {}
    # 單層迴圈，時間複雜度為 O(n)
    for i in range(len(nums)):
        if target - nums[i] in dic:
            return [dic[target - nums[i]], i]
        dic[nums[i]] = i
    return []
```

=== "C++"

```cpp title="two_sum.cpp"
/* 方法二：輔助雜湊表 */
vector<int> twoSumHashTable(vector<int> &nums, int target) {
    int size = nums.size();
    // 輔助雜湊表，空間複雜度為 O(n)
    unordered_map<int, int> dic;
    // 單層迴圈，時間複雜度為 O(n)
    for (int i = 0; i < size; i++) {
        if (dic.find(target - nums[i]) != dic.end()) {
            return {dic[target - nums[i]], i};
        }
        dic.emplace(nums[i], i);
    }
    return {};
}
```

=== "Java"

```java title="two_sum.java"
/* 方法二：輔助雜湊表 */
int[] twoSumHashTable(int[] nums, int target) {
    int size = nums.length;
    // 輔助雜湊表，空間複雜度為 O(n)
    Map<Integer, Integer> dic = new HashMap<>();
    // 單層迴圈，時間複雜度為 O(n)
    for (int i = 0; i < size; i++) {
        if (dic.containsKey(target - nums[i])) {
            return new int[] { dic.get(target - nums[i]), i };
        }
        dic.put(nums[i], i);
    }
    return new int[0];
}
```

=== "C#"

```csharp title="two_sum.cs"
/* 方法二：輔助雜湊表 */
int[] TwoSumHashTable(int[] nums, int target) {
    int size = nums.Length;
    // 輔助雜湊表，空間複雜度為 O(n)
    Dictionary<int, int> dic = [];
    // 單層迴圈，時間複雜度為 O(n)
    for (int i = 0; i < size; i++) {
        if (dic.ContainsKey(target - nums[i])) {
            return [dic[target - nums[i]], i];
        }
        dic.Add(nums[i], i);
    }
    return [];
}
```

=== "Go"

```go title="two_sum.go"
/* 方法二：輔助雜湊表 */
func twoSumHashTable(nums []int, target int) []int {
    // 輔助雜湊表，空間複雜度為 O(n)
    hashTable := map[int]int{}
    // 單層迴圈，時間複雜度為 O(n)
    for idx, val := range nums {
        if preIdx, ok := hashTable[target-val]; ok {
            return []int{preIdx, idx}
        }
        hashTable[val] = idx
    }
    return nil
}
```

=== "Swift"

```swift title="two_sum.swift"
/* 方法二：輔助雜湊表 */
func twoSumHashTable(nums: [Int], target: Int) -> [Int] {
    // 輔助雜湊表，空間複雜度為 O(n)
    var dic: [Int: Int] = [:]
    // 單層迴圈，時間複雜度為 O(n)
    for i in nums.indices {
        if let j = dic[target - nums[i]] {
            return [j, i]
        }
        dic[nums[i]] = i
    }
    return [0]
}
```

=== "JS"

```javascript title="two_sum.js"
/* 方法二：輔助雜湊表 */
function twoSumHashTable(nums, target) {
    // 輔助雜湊表，空間複雜度為 O(n)
    let m = {};
    // 單層迴圈，時間複雜度為 O(n)
    for (let i = 0; i < nums.length; i++) {
        if (m[target - nums[i]] !== undefined) {
            return [m[target - nums[i]], i];
        } else {
            m[nums[i]] = i;
        }
    }
    return [];
}
```

=== "TS"

```typescript title="two_sum.ts"
/* 方法二：輔助雜湊表 */
function twoSumHashTable(nums: number[], target: number): number[] {
    // 輔助雜湊表，空間複雜度為 O(n)
    let m: Map<number, number> = new Map();
    // 單層迴圈，時間複雜度為 O(n)
    for (let i = 0; i < nums.length; i++) {
        let index = m.get(target - nums[i]);
        if (index !== undefined) {
            return [index, i];
        } else {
            m.set(nums[i], i);
        }
    }
    return [];
}
```

=== "Dart"

```dart title="two_sum.dart"
/* 方法二： 輔助雜湊表 */
List<int> twoSumHashTable(List<int> nums, int target) {
  int size = nums.length;
  // 輔助雜湊表，空間複雜度為 O(n)
  Map<int, int> dic = HashMap();
  // 單層迴圈，時間複雜度為 O(n)
  for (var i = 0; i < size; i++) {
    if (dic.containsKey(target - nums[i])) {
      return [dic[target - nums[i]]!, i];
    }
    dic.putIfAbsent(nums[i], () => i);
  }
  return [0];
}
```

=== "Rust"

```rust title="two_sum.rs"
/* 方法二：輔助雜湊表 */
pub fn two_sum_hash_table(nums: &Vec<i32>, target: i32) -> Option<Vec<i32>> {
    // 輔助雜湊表，空間複雜度為 O(n)
    let mut dic = HashMap::new();
    // 單層迴圈，時間複雜度為 O(n)
    for (i, num) in nums.iter().enumerate() {
        match dic.get(&(target - num)) {
            Some(v) => return Some(vec![*v as i32, i as i32]),
            None => dic.insert(num, i as i32),
        };
    }
    None
}
```

=== "C"

```c title="two_sum.c"
/* 雜湊表 */
typedef struct {
    int key;
    int val;
    UT_hash_handle hh; // 基於 uthash.h 實現
} HashTable;

/* 雜湊表查詢 */
HashTable *find(HashTable *h, int key) {
    HashTable *tmp;
    HASH_FIND_INT(h, &key, tmp);
    return tmp;
}

/* 雜湊表元素插入 */
void insert(HashTable *h, int key, int val) {
    HashTable *t = find(h, key);
    if (t == NULL) {
        HashTable *tmp = malloc(sizeof(HashTable));
        tmp->key = key, tmp->val = val;
        HASH_ADD_INT(h, key, tmp);
    } else {
        t->val = val;
    }
}

/* 方法二：輔助雜湊表 */
int *twoSumHashTable(int *nums, int numsSize, int target, int *returnSize) {
    HashTable *hashtable = NULL;
    for (int i = 0; i < numsSize; i++) {
        HashTable *t = find(hashtable, target - nums[i]);
        if (t != NULL) {
            int *res = malloc(sizeof(int) * 2);
            res[0] = t->val, res[1] = i;
            *returnSize = 2;
            return res;
        }
        insert(hashtable, nums[i], i);
    }
    *returnSize = 0;
    return NULL;
}
```

=== "Kotlin"

```kotlin title="two_sum.kt"
/* 方法二：輔助雜湊表 */
fun twoSumHashTable(nums: IntArray, target: Int): IntArray {
    val size = nums.size
    // 輔助雜湊表，空間複雜度為 O(n)
    val dic = HashMap<Int, Int>()
    // 單層迴圈，時間複雜度為 O(n)
    for (i in 0..<size) {
        if (dic.containsKey(target - nums[i])) {
            return intArrayOf(dic[target - nums[i]]!!, i)
        }
        dic[nums[i]] = i
    }
    return IntArray(0)
}
```

=== "Ruby"

```ruby title="two_sum.rb"
### 方法二：輔助雜湊表 ###
def two_sum_hash_table(nums, target)
  # 輔助雜湊表，空間複雜度為 O(n)
  dic = {}
  # 單層迴圈，時間複雜度為 O(n)
  for i in 0...nums.length
    return [dic[target - nums[i]], i] if dic.has_key?(target - nums[i])

    dic[nums[i]] = i
  end

  []
end
```

=== "Zig"

```zig title="two_sum.zig"
// 方法二：輔助雜湊表
fn twoSumHashTable(nums: []i32, target: i32) !?[2]i32 {
    var size: usize = nums.len;
    // 輔助雜湊表，空間複雜度為 O(n)
    var dic = std.AutoHashMap(i32, i32).init(std.heap.page_allocator);
    defer dic.deinit();
    var i: usize = 0;
    // 單層迴圈，時間複雜度為 O(n)
    while (i < size) : (i += 1) {
        if (dic.contains(target - nums[i])) {
            return [_]i32{dic.get(target - nums[i]).?, @intCast(i)};
        }
        try dic.put(nums[i], @intCast(i));
    }
    return null;
}
```

??? pythontutor "視覺化執行"

<div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20two_sum_hash_table%28nums%3A%20list%5Bint%5D%2C%20target%3A%20int%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%22%22%22%E6%96%B9%E6%B3%95%E4%BA%8C%EF%BC%9A%E8%BC%94%E5%8A%A9%E9%9B%9C%E6%B9%8A%E8%A1%A8%22%22%22%0A%20%20%20%20%23%20%E8%BC%94%E5%8A%A9%E9%9B%9C%E6%B9%8A%E8%A1%A8%EF%BC%8C%E7%A9%BA%E9%96%93%E8%A4%87%E9%9B%9C%E5%BA%A6%E7%82%BA%20O%28n%29%0A%20%20%20%20dic%20%3D%20%7B%7D%0A%20%20%20%20%23%20%E5%96%AE%E5%B1%A4%E8%BF%B4%E5%9C%88%EF%BC%8C%E6%99%82%E9%96%93%E8%A4%87%E9%9B%9C%E5%BA%A6%E7%82%BA%20O%28n%29%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20if%20target%20-%20nums%5Bi%5D%20in%20dic%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20%5Bdic%5Btarget%20-%20nums%5Bi%5D%5D%2C%20i%5D%0A%20%20%20%20%20%20%20%20dic%5Bnums%5Bi%5D%5D%20%3D%20i%0A%20%20%20%20return%20%5B%5D%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%5B2%2C%207%2C%2011%2C%2015%5D%0A%20%20%20%20target%20%3D%2013%0A%20%20%20%20res%20%3D%20two_sum_hash_table%28nums%2C%20target%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20two_sum_hash_table%28nums%3A%20list%5Bint%5D%2C%20target%3A%20int%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%22%22%22%E6%96%B9%E6%B3%95%E4%BA%8C%EF%BC%9A%E8%BC%94%E5%8A%A9%E9%9B%9C%E6%B9%8A%E8%A1%A8%22%22%22%0A%20%20%20%20%23%20%E8%BC%94%E5%8A%A9%E9%9B%9C%E6%B9%8A%E8%A1%A8%EF%BC%8C%E7%A9%BA%E9%96%93%E8%A4%87%E9%9B%9C%E5%BA%A6%E7%82%BA%20O%28n%29%0A%20%20%20%20dic%20%3D%20%7B%7D%0A%20%20%20%20%23%20%E5%96%AE%E5%B1%A4%E8%BF%B4%E5%9C%88%EF%BC%8C%E6%99%82%E9%96%93%E8%A4%87%E9%9B%9C%E5%BA%A6%E7%82%BA%20O%28n%29%0A%20%20%20%20for%20i%20in%20range%28len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20if%20target%20-%20nums%5Bi%5D%20in%20dic%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20%5Bdic%5Btarget%20-%20nums%5Bi%5D%5D%2C%20i%5D%0A%20%20%20%20%20%20%20%20dic%5Bnums%5Bi%5D%5D%20%3D%20i%0A%20%20%20%20return%20%5B%5D%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%5B2%2C%207%2C%2011%2C%2015%5D%0A%20%20%20%20target%20%3D%2013%0A%20%20%20%20res%20%3D%20two_sum_hash_table%28nums%2C%20target%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全螢幕觀看 ></a></div>

此方法透過雜湊查詢將時間複雜度從 O(n^2) 降至 O(n) ，大幅提升執行效率。

由於需要維護一個額外的雜湊表，因此空間複雜度為 O(n) 。儘管如此，該方法的整體時空效率更為均衡，因此它是本題的最優解法。