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62 lines
1.9 KiB
62 lines
1.9 KiB
/*
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* File: subset_sum_ii.rs
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* Created Time: 2023-07-09
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* Author: codingonion (coderonion@gmail.com)
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*/
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/* 回溯演算法：子集和 II */
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fn backtrack(
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mut state: Vec<i32>,
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target: i32,
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choices: &[i32],
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start: usize,
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res: &mut Vec<Vec<i32>>,
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) {
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// 子集和等於 target 時，記錄解
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if target == 0 {
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res.push(state);
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return;
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}
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// 走訪所有選擇
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// 剪枝二：從 start 開始走訪，避免生成重複子集
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// 剪枝三：從 start 開始走訪，避免重複選擇同一元素
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for i in start..choices.len() {
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// 剪枝一：若子集和超過 target ，則直接結束迴圈
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// 這是因為陣列已排序，後邊元素更大，子集和一定超過 target
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if target - choices[i] < 0 {
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break;
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}
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// 剪枝四：如果該元素與左邊元素相等，說明該搜尋分支重複，直接跳過
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if i > start && choices[i] == choices[i - 1] {
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continue;
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}
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// 嘗試：做出選擇，更新 target, start
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state.push(choices[i]);
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// 進行下一輪選擇
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backtrack(state.clone(), target - choices[i], choices, i + 1, res);
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// 回退：撤銷選擇，恢復到之前的狀態
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state.pop();
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}
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}
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/* 求解子集和 II */
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fn subset_sum_ii(nums: &mut [i32], target: i32) -> Vec<Vec<i32>> {
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let state = Vec::new(); // 狀態（子集）
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nums.sort(); // 對 nums 進行排序
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let start = 0; // 走訪起始點
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let mut res = Vec::new(); // 結果串列（子集串列）
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backtrack(state, target, nums, start, &mut res);
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res
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}
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/* Driver Code */
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pub fn main() {
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let mut nums = [4, 4, 5];
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let target = 9;
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let res = subset_sum_ii(&mut nums, target);
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println!("輸入陣列 nums = {:?}, target = {}", &nums, target);
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println!("所有和等於 {} 的子集 res = {:?}", target, &res);
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}
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