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782 lines
25 KiB
Markdown
782 lines
25 KiB
Markdown
<p align="center">
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<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
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<img src="../pics/训练营.png" width="1000"/>
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</a>
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<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
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> 这篇可以说是全网把组合问题如何去重，讲的最清晰的了！
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# 40.组合总和II
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[力扣题目链接](https://leetcode.cn/problems/combination-sum-ii/)
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给定一个数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
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candidates 中的每个数字在每个组合中只能使用一次。
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说明：
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所有数字（包括目标数）都是正整数。解集不能包含重复的组合。
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* 示例 1:
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* 输入: candidates = [10,1,2,7,6,1,5], target = 8,
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* 所求解集为:
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```
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[
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[1, 7],
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[1, 2, 5],
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[2, 6],
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[1, 1, 6]
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]
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```
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* 示例 2:
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* 输入: candidates = [2,5,2,1,2], target = 5,
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* 所求解集为:
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```
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[
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[1,2,2],
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[5]
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]
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```
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## 算法公开课
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[回溯算法中的去重，树层去重树枝去重，你弄清楚了没？| LeetCode:40.组合总和II](https://www.bilibili.com/video/BV12V4y1V73A)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。
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## 思路
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这道题目和[39.组合总和](https://programmercarl.com/0039.组合总和.html)如下区别：
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1. 本题candidates 中的每个数字在每个组合中只能使用一次。
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2. 本题数组candidates的元素是有重复的，而[39.组合总和](https://programmercarl.com/0039.组合总和.html)是无重复元素的数组candidates
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最后本题和[39.组合总和](https://programmercarl.com/0039.组合总和.html)要求一样，解集不能包含重复的组合。
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**本题的难点在于区别2中：集合（数组candidates）有重复元素，但还不能有重复的组合**。
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一些同学可能想了：我把所有组合求出来，再用set或者map去重，这么做很容易超时！
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所以要在搜索的过程中就去掉重复组合。
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很多同学在去重的问题上想不明白，其实很多题解也没有讲清楚，反正代码是能过的，感觉是那么回事，稀里糊涂的先把题目过了。
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这个去重为什么很难理解呢，**所谓去重，其实就是使用过的元素不能重复选取。** 这么一说好像很简单！
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都知道组合问题可以抽象为树形结构，那么“使用过”在这个树形结构上是有两个维度的，一个维度是同一树枝上使用过，一个维度是同一树层上使用过。**没有理解这两个层面上的“使用过” 是造成大家没有彻底理解去重的根本原因。**
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那么问题来了，我们是要同一树层上使用过，还是同一树枝上使用过呢？
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回看一下题目，元素在同一个组合内是可以重复的，怎么重复都没事，但两个组合不能相同。
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**所以我们要去重的是同一树层上的“使用过”，同一树枝上的都是一个组合里的元素，不用去重**。
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为了理解去重我们来举一个例子，candidates = [1, 1, 2], target = 3，（方便起见candidates已经排序了）
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**强调一下，树层去重的话，需要对数组排序！**
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选择过程树形结构如图所示：
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可以看到图中，每个节点相对于 [39.组合总和](https://mp.weixin.qq.com/s/FLg8G6EjVcxBjwCbzpACPw)我多加了used数组，这个used数组下面会重点介绍。
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### 回溯三部曲
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* **递归函数参数**
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与[39.组合总和](https://programmercarl.com/0039.组合总和.html)套路相同，此题还需要加一个bool型数组used，用来记录同一树枝上的元素是否使用过。
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这个集合去重的重任就是used来完成的。
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代码如下：
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```CPP
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vector<vector<int>> result; // 存放组合集合
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vector<int> path; // 符合条件的组合
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void backtracking(vector<int>& candidates, int target, int sum, int startIndex, vector<bool>& used) {
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```
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* **递归终止条件**
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与[39.组合总和](https://programmercarl.com/0039.组合总和.html)相同，终止条件为 `sum > target` 和 `sum == target`。
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代码如下：
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```CPP
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if (sum > target) { // 这个条件其实可以省略
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return;
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}
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if (sum == target) {
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result.push_back(path);
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return;
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}
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```
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`sum > target` 这个条件其实可以省略，因为在递归单层遍历的时候，会有剪枝的操作，下面会介绍到。
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* **单层搜索的逻辑**
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这里与[39.组合总和](https://programmercarl.com/0039.组合总和.html)最大的不同就是要去重了。
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前面我们提到：要去重的是“同一树层上的使用过”，如何判断同一树层上元素（相同的元素）是否使用过了呢。
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**如果`candidates[i] == candidates[i - 1]` 并且 `used[i - 1] == false`，就说明：前一个树枝，使用了candidates[i - 1]，也就是说同一树层使用过candidates[i - 1]**。
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此时for循环里就应该做continue的操作。
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这块比较抽象，如图：
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我在图中将used的变化用橘黄色标注上，可以看出在candidates[i] == candidates[i - 1]相同的情况下：
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* used[i - 1] == true，说明同一树枝candidates[i - 1]使用过
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* used[i - 1] == false，说明同一树层candidates[i - 1]使用过
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可能有的录友想，为什么 used[i - 1] == false 就是同一树层呢，因为同一树层，used[i - 1] == false 才能表示，当前取的 candidates[i] 是从 candidates[i - 1] 回溯而来的。
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而 used[i - 1] == true，说明是进入下一层递归，去下一个数，所以是树枝上，如图所示：
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**这块去重的逻辑很抽象，网上搜的题解基本没有能讲清楚的，如果大家之前思考过这个问题或者刷过这道题目，看到这里一定会感觉通透了很多！**
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那么单层搜索的逻辑代码如下：
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```CPP
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for (int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++) {
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// used[i - 1] == true，说明同一树枝candidates[i - 1]使用过
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// used[i - 1] == false，说明同一树层candidates[i - 1]使用过
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// 要对同一树层使用过的元素进行跳过
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if (i > 0 && candidates[i] == candidates[i - 1] && used[i - 1] == false) {
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continue;
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}
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sum += candidates[i];
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path.push_back(candidates[i]);
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used[i] = true;
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backtracking(candidates, target, sum, i + 1, used); // 和39.组合总和的区别1：这里是i+1，每个数字在每个组合中只能使用一次
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used[i] = false;
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sum -= candidates[i];
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path.pop_back();
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}
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```
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**注意sum + candidates[i] <= target为剪枝操作，在[39.组合总和](https://mp.weixin.qq.com/s/FLg8G6EjVcxBjwCbzpACPw)有讲解过！**
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回溯三部曲分析完了，整体C++代码如下：
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```CPP
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class Solution {
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private:
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vector<vector<int>> result;
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vector<int> path;
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void backtracking(vector<int>& candidates, int target, int sum, int startIndex, vector<bool>& used) {
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if (sum == target) {
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result.push_back(path);
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return;
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}
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for (int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++) {
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// used[i - 1] == true，说明同一树枝candidates[i - 1]使用过
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// used[i - 1] == false，说明同一树层candidates[i - 1]使用过
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// 要对同一树层使用过的元素进行跳过
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if (i > 0 && candidates[i] == candidates[i - 1] && used[i - 1] == false) {
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continue;
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}
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sum += candidates[i];
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path.push_back(candidates[i]);
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used[i] = true;
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backtracking(candidates, target, sum, i + 1, used); // 和39.组合总和的区别1，这里是i+1，每个数字在每个组合中只能使用一次
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used[i] = false;
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sum -= candidates[i];
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path.pop_back();
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}
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}
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public:
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vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
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vector<bool> used(candidates.size(), false);
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path.clear();
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result.clear();
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// 首先把给candidates排序，让其相同的元素都挨在一起。
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sort(candidates.begin(), candidates.end());
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backtracking(candidates, target, 0, 0, used);
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return result;
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}
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};
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```
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* 时间复杂度: O(n * 2^n)
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* 空间复杂度: O(n)
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### 补充
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这里直接用startIndex来去重也是可以的， 就不用used数组了。
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```CPP
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class Solution {
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private:
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vector<vector<int>> result;
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vector<int> path;
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void backtracking(vector<int>& candidates, int target, int sum, int startIndex) {
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if (sum == target) {
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result.push_back(path);
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return;
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}
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for (int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++) {
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// 要对同一树层使用过的元素进行跳过
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if (i > startIndex && candidates[i] == candidates[i - 1]) {
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continue;
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}
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sum += candidates[i];
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path.push_back(candidates[i]);
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backtracking(candidates, target, sum, i + 1); // 和39.组合总和的区别1，这里是i+1，每个数字在每个组合中只能使用一次
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sum -= candidates[i];
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path.pop_back();
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}
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}
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public:
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vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
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path.clear();
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result.clear();
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// 首先把给candidates排序，让其相同的元素都挨在一起。
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sort(candidates.begin(), candidates.end());
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backtracking(candidates, target, 0, 0);
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return result;
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}
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};
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```
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## 总结
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本题同样是求组合总和，但就是因为其数组candidates有重复元素，而要求不能有重复的组合，所以相对于[39.组合总和](https://programmercarl.com/0039.组合总和.html)难度提升了不少。
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**关键是去重的逻辑，代码很简单，网上一搜一大把，但几乎没有能把这块代码含义讲明白的，基本都是给出代码，然后说这就是去重了，究竟怎么个去重法也是模棱两可**。
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所以Carl有必要把去重的这块彻彻底底的给大家讲清楚，**就连“树层去重”和“树枝去重”都是我自创的词汇，希望对大家理解有帮助！**
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## 其他语言版本
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### Java
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**使用标记数组**
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```Java
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class Solution {
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LinkedList<Integer> path = new LinkedList<>();
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List<List<Integer>> ans = new ArrayList<>();
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boolean[] used;
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int sum = 0;
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public List<List<Integer>> combinationSum2(int[] candidates, int target) {
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used = new boolean[candidates.length];
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// 加标志数组，用来辅助判断同层节点是否已经遍历
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Arrays.fill(used, false);
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// 为了将重复的数字都放到一起，所以先进行排序
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Arrays.sort(candidates);
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backTracking(candidates, target, 0);
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return ans;
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}
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private void backTracking(int[] candidates, int target, int startIndex) {
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if (sum == target) {
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ans.add(new ArrayList(path));
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}
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for (int i = startIndex; i < candidates.length; i++) {
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if (sum + candidates[i] > target) {
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break;
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}
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// 出现重复节点，同层的第一个节点已经被访问过，所以直接跳过
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if (i > 0 && candidates[i] == candidates[i - 1] && !used[i - 1]) {
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continue;
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}
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used[i] = true;
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sum += candidates[i];
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path.add(candidates[i]);
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// 每个节点仅能选择一次，所以从下一位开始
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backTracking(candidates, target, i + 1);
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used[i] = false;
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sum -= candidates[i];
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path.removeLast();
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}
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}
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}
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```
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**不使用标记数组**
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```Java
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class Solution {
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List<List<Integer>> res = new ArrayList<>();
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LinkedList<Integer> path = new LinkedList<>();
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int sum = 0;
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public List<List<Integer>> combinationSum2( int[] candidates, int target ) {
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//为了将重复的数字都放到一起，所以先进行排序
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Arrays.sort( candidates );
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backTracking( candidates, target, 0 );
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return res;
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}
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private void backTracking( int[] candidates, int target, int start ) {
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if ( sum == target ) {
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res.add( new ArrayList<>( path ) );
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return;
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}
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for ( int i = start; i < candidates.length && sum + candidates[i] <= target; i++ ) {
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//正确剔除重复解的办法
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//跳过同一树层使用过的元素
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if ( i > start && candidates[i] == candidates[i - 1] ) {
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continue;
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}
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sum += candidates[i];
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path.add( candidates[i] );
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// i+1 代表当前组内元素只选取一次
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backTracking( candidates, target, i + 1 );
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int temp = path.getLast();
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sum -= temp;
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path.removeLast();
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}
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}
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}
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```
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### Python
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回溯
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```python
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class Solution:
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def backtracking(self, candidates, target, total, startIndex, path, result):
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if total == target:
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result.append(path[:])
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return
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for i in range(startIndex, len(candidates)):
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if i > startIndex and candidates[i] == candidates[i - 1]:
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continue
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if total + candidates[i] > target:
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break
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total += candidates[i]
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path.append(candidates[i])
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self.backtracking(candidates, target, total, i + 1, path, result)
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total -= candidates[i]
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path.pop()
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def combinationSum2(self, candidates, target):
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result = []
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candidates.sort()
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self.backtracking(candidates, target, 0, 0, [], result)
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return result
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```
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回溯 使用used
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```python
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class Solution:
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def backtracking(self, candidates, target, total, startIndex, used, path, result):
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if total == target:
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result.append(path[:])
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return
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for i in range(startIndex, len(candidates)):
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# 对于相同的数字，只选择第一个未被使用的数字，跳过其他相同数字
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if i > startIndex and candidates[i] == candidates[i - 1] and not used[i - 1]:
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continue
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if total + candidates[i] > target:
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break
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total += candidates[i]
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path.append(candidates[i])
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used[i] = True
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self.backtracking(candidates, target, total, i + 1, used, path, result)
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used[i] = False
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total -= candidates[i]
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path.pop()
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def combinationSum2(self, candidates, target):
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used = [False] * len(candidates)
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result = []
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candidates.sort()
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self.backtracking(candidates, target, 0, 0, used, [], result)
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return result
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```
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回溯优化
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```python
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class Solution:
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def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
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candidates.sort()
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results = []
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self.combinationSumHelper(candidates, target, 0, [], results)
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return results
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def combinationSumHelper(self, candidates, target, index, path, results):
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if target == 0:
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results.append(path[:])
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return
|
||
for i in range(index, len(candidates)):
|
||
if i > index and candidates[i] == candidates[i - 1]:
|
||
continue
|
||
if candidates[i] > target:
|
||
break
|
||
path.append(candidates[i])
|
||
self.combinationSumHelper(candidates, target - candidates[i], i + 1, path, results)
|
||
path.pop()
|
||
```
|
||
### Go
|
||
主要在于如何在回溯中去重
|
||
|
||
**使用used数组**
|
||
```go
|
||
var (
|
||
res [][]int
|
||
path []int
|
||
used []bool
|
||
)
|
||
func combinationSum2(candidates []int, target int) [][]int {
|
||
res, path = make([][]int, 0), make([]int, 0, len(candidates))
|
||
used = make([]bool, len(candidates))
|
||
sort.Ints(candidates) // 排序，为剪枝做准备
|
||
dfs(candidates, 0, target)
|
||
return res
|
||
}
|
||
|
||
func dfs(candidates []int, start int, target int) {
|
||
if target == 0 { // target 不断减小，如果为0说明达到了目标值
|
||
tmp := make([]int, len(path))
|
||
copy(tmp, path)
|
||
res = append(res, tmp)
|
||
return
|
||
}
|
||
for i := start; i < len(candidates); i++ {
|
||
if candidates[i] > target { // 剪枝，提前返回
|
||
break
|
||
}
|
||
// used[i - 1] == true，说明同一树枝candidates[i - 1]使用过
|
||
// used[i - 1] == false，说明同一树层candidates[i - 1]使用过
|
||
if i > 0 && candidates[i] == candidates[i-1] && used[i-1] == false {
|
||
continue
|
||
}
|
||
path = append(path, candidates[i])
|
||
used[i] = true
|
||
dfs(candidates, i+1, target - candidates[i])
|
||
used[i] = false
|
||
path = path[:len(path) - 1]
|
||
}
|
||
}
|
||
```
|
||
**不使用used数组**
|
||
```go
|
||
var (
|
||
res [][]int
|
||
path []int
|
||
)
|
||
func combinationSum2(candidates []int, target int) [][]int {
|
||
res, path = make([][]int, 0), make([]int, 0, len(candidates))
|
||
sort.Ints(candidates) // 排序，为剪枝做准备
|
||
dfs(candidates, 0, target)
|
||
return res
|
||
}
|
||
|
||
func dfs(candidates []int, start int, target int) {
|
||
if target == 0 { // target 不断减小，如果为0说明达到了目标值
|
||
tmp := make([]int, len(path))
|
||
copy(tmp, path)
|
||
res = append(res, tmp)
|
||
return
|
||
}
|
||
for i := start; i < len(candidates); i++ {
|
||
if candidates[i] > target { // 剪枝，提前返回
|
||
break
|
||
}
|
||
// i != start 限制了这不对深度遍历到达的此值去重
|
||
if i != start && candidates[i] == candidates[i-1] { // 去重
|
||
continue
|
||
}
|
||
path = append(path, candidates[i])
|
||
dfs(candidates, i+1, target - candidates[i])
|
||
path = path[:len(path) - 1]
|
||
}
|
||
}
|
||
```
|
||
### JavaScript
|
||
|
||
```js
|
||
/**
|
||
* @param {number[]} candidates
|
||
* @param {number} target
|
||
* @return {number[][]}
|
||
*/
|
||
var combinationSum2 = function(candidates, target) {
|
||
const res = []; path = [], len = candidates.length;
|
||
candidates.sort((a,b)=>a-b);
|
||
backtracking(0, 0);
|
||
return res;
|
||
function backtracking(sum, i) {
|
||
if (sum === target) {
|
||
res.push(Array.from(path));
|
||
return;
|
||
}
|
||
for(let j = i; j < len; j++) {
|
||
const n = candidates[j];
|
||
if(j > i && candidates[j] === candidates[j-1]){
|
||
//若当前元素和前一个元素相等
|
||
//则本次循环结束，防止出现重复组合
|
||
continue;
|
||
}
|
||
//如果当前元素值大于目标值-总和的值
|
||
//由于数组已排序，那么该元素之后的元素必定不满足条件
|
||
//直接终止当前层的递归
|
||
if(n > target - sum) break;
|
||
path.push(n);
|
||
sum += n;
|
||
backtracking(sum, j + 1);
|
||
path.pop();
|
||
sum -= n;
|
||
}
|
||
}
|
||
};
|
||
```
|
||
**使用used去重**
|
||
|
||
```js
|
||
var combinationSum2 = function(candidates, target) {
|
||
let res = [];
|
||
let path = [];
|
||
let total = 0;
|
||
const len = candidates.length;
|
||
candidates.sort((a, b) => a - b);
|
||
let used = new Array(len).fill(false);
|
||
const backtracking = (startIndex) => {
|
||
if (total === target) {
|
||
res.push([...path]);
|
||
return;
|
||
}
|
||
for(let i = startIndex; i < len && total < target; i++) {
|
||
const cur = candidates[i];
|
||
if (cur > target - total || (i > 0 && cur === candidates[i - 1] && !used[i - 1])) continue;
|
||
path.push(cur);
|
||
total += cur;
|
||
used[i] = true;
|
||
backtracking(i + 1);
|
||
path.pop();
|
||
total -= cur;
|
||
used[i] = false;
|
||
}
|
||
}
|
||
backtracking(0);
|
||
return res;
|
||
};
|
||
```
|
||
|
||
### TypeScript
|
||
|
||
```typescript
|
||
function combinationSum2(candidates: number[], target: number): number[][] {
|
||
candidates.sort((a, b) => a - b);
|
||
const resArr: number[][] = [];
|
||
function backTracking(
|
||
candidates: number[], target: number,
|
||
curSum: number, startIndex: number, route: number[]
|
||
) {
|
||
if (curSum > target) return;
|
||
if (curSum === target) {
|
||
resArr.push(route.slice());
|
||
return;
|
||
}
|
||
for (let i = startIndex, length = candidates.length; i < length; i++) {
|
||
if (i > startIndex && candidates[i] === candidates[i - 1]) {
|
||
continue;
|
||
}
|
||
let tempVal: number = candidates[i];
|
||
route.push(tempVal);
|
||
backTracking(candidates, target, curSum + tempVal, i + 1, route);
|
||
route.pop();
|
||
|
||
}
|
||
}
|
||
backTracking(candidates, target, 0, 0, []);
|
||
return resArr;
|
||
};
|
||
```
|
||
|
||
### Rust
|
||
|
||
```Rust
|
||
impl Solution {
|
||
pub fn backtracking(result: &mut Vec<Vec<i32>>, path: &mut Vec<i32>, candidates: &Vec<i32>, target: i32, mut sum: i32, start_index: usize, used: &mut Vec<bool>) {
|
||
if sum == target {
|
||
result.push(path.to_vec());
|
||
return;
|
||
}
|
||
for i in start_index..candidates.len() {
|
||
if sum + candidates[i] <= target {
|
||
if i > 0 && candidates[i] == candidates[i - 1] && used[i - 1] == false { continue; }
|
||
sum += candidates[i];
|
||
path.push(candidates[i]);
|
||
used[i] = true;
|
||
Self::backtracking(result, path, candidates, target, sum, i + 1, used);
|
||
used[i] = false;
|
||
sum -= candidates[i];
|
||
path.pop();
|
||
}
|
||
}
|
||
}
|
||
|
||
pub fn combination_sum2(candidates: Vec<i32>, target: i32) -> Vec<Vec<i32>> {
|
||
let mut result: Vec<Vec<i32>> = Vec::new();
|
||
let mut path: Vec<i32> = Vec::new();
|
||
let mut used: Vec<bool> = vec![false; candidates.len()];
|
||
let mut candidates = candidates;
|
||
candidates.sort();
|
||
Self::backtracking(&mut result, &mut path, &candidates, target, 0, 0, &mut used);
|
||
result
|
||
}
|
||
}
|
||
```
|
||
|
||
### C
|
||
|
||
```c
|
||
int* path;
|
||
int pathTop;
|
||
int** ans;
|
||
int ansTop;
|
||
//记录ans中每一个一维数组的大小
|
||
int* length;
|
||
int cmp(const void* a1, const void* a2) {
|
||
return *((int*)a1) - *((int*)a2);
|
||
}
|
||
|
||
void backTracking(int* candidates, int candidatesSize, int target, int sum, int startIndex) {
|
||
if(sum >= target) {
|
||
//若sum等于target，复制当前path进入
|
||
if(sum == target) {
|
||
int* tempPath = (int*)malloc(sizeof(int) * pathTop);
|
||
int j;
|
||
for(j = 0; j < pathTop; j++) {
|
||
tempPath[j] = path[j];
|
||
}
|
||
length[ansTop] = pathTop;
|
||
ans[ansTop++] = tempPath;
|
||
}
|
||
return ;
|
||
}
|
||
|
||
int i;
|
||
for(i = startIndex; i < candidatesSize; i++) {
|
||
//对同一层树中使用过的元素跳过
|
||
if(i > startIndex && candidates[i] == candidates[i-1])
|
||
continue;
|
||
path[pathTop++] = candidates[i];
|
||
sum += candidates[i];
|
||
backTracking(candidates, candidatesSize, target, sum, i + 1);
|
||
//回溯
|
||
sum -= candidates[i];
|
||
pathTop--;
|
||
}
|
||
}
|
||
|
||
int** combinationSum2(int* candidates, int candidatesSize, int target, int* returnSize, int** returnColumnSizes){
|
||
path = (int*)malloc(sizeof(int) * 50);
|
||
ans = (int**)malloc(sizeof(int*) * 100);
|
||
length = (int*)malloc(sizeof(int) * 100);
|
||
pathTop = ansTop = 0;
|
||
//快速排序candidates，让相同元素挨到一起
|
||
qsort(candidates, candidatesSize, sizeof(int), cmp);
|
||
|
||
backTracking(candidates, candidatesSize, target, 0, 0);
|
||
|
||
*returnSize = ansTop;
|
||
*returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
|
||
int i;
|
||
for(i = 0; i < ansTop; i++) {
|
||
(*returnColumnSizes)[i] = length[i];
|
||
}
|
||
return ans;
|
||
}
|
||
```
|
||
|
||
### Swift
|
||
|
||
```swift
|
||
func combinationSum2(_ candidates: [Int], _ target: Int) -> [[Int]] {
|
||
// 为了方便去重复，先对集合排序
|
||
let candidates = candidates.sorted()
|
||
var result = [[Int]]()
|
||
var path = [Int]()
|
||
func backtracking(sum: Int, startIndex: Int) {
|
||
// 终止条件
|
||
if sum == target {
|
||
result.append(path)
|
||
return
|
||
}
|
||
|
||
let end = candidates.count
|
||
guard startIndex < end else { return }
|
||
for i in startIndex ..< end {
|
||
if i > startIndex, candidates[i] == candidates[i - 1] { continue } // 去重复
|
||
let sum = sum + candidates[i] // 使用局部变量隐藏回溯
|
||
if sum > target { continue } // 剪枝
|
||
|
||
path.append(candidates[i]) // 处理
|
||
backtracking(sum: sum, startIndex: i + 1) // i+1避免重复访问
|
||
path.removeLast() // 回溯
|
||
}
|
||
}
|
||
backtracking(sum: 0, startIndex: 0)
|
||
return result
|
||
}
|
||
```
|
||
|
||
|
||
### Scala
|
||
|
||
```scala
|
||
object Solution {
|
||
import scala.collection.mutable
|
||
def combinationSum2(candidates: Array[Int], target: Int): List[List[Int]] = {
|
||
var res = mutable.ListBuffer[List[Int]]()
|
||
var path = mutable.ListBuffer[Int]()
|
||
var candidate = candidates.sorted
|
||
|
||
def backtracking(sum: Int, startIndex: Int): Unit = {
|
||
if (sum == target) {
|
||
res.append(path.toList)
|
||
return
|
||
}
|
||
|
||
for (i <- startIndex until candidate.size if sum + candidate(i) <= target) {
|
||
if (!(i > startIndex && candidate(i) == candidate(i - 1))) {
|
||
path.append(candidate(i))
|
||
backtracking(sum + candidate(i), i + 1)
|
||
path = path.take(path.size - 1)
|
||
}
|
||
}
|
||
}
|
||
|
||
backtracking(0, 0)
|
||
res.toList
|
||
}
|
||
}
|
||
```
|
||
|
||
<p align="center">
|
||
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
|
||
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
|
||
</a>
|
||
|