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607 lines
19 KiB
Markdown
607 lines
19 KiB
Markdown
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<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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# 39. 组合总和
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[力扣题目链接](https://leetcode.cn/problems/combination-sum/)
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给定一个无重复元素的数组 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 中的数字可以无限制重复被选取。
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说明：
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* 所有数字（包括 target）都是正整数。
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* 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 = [2,3,6,7], target = 7,
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* 所求解集为：
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[
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[7],
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[2,2,3]
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]
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示例 2：
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* 输入：candidates = [2,3,5], target = 8,
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* 所求解集为：
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[
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[2,2,2,2],
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[2,3,3],
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[3,5]
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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：[Leetcode:39. 组合总和讲解](https://www.bilibili.com/video/BV1KT4y1M7HJ)，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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题目中的**无限制重复被选取，吓得我赶紧想想 出现0 可咋办**，然后看到下面提示：1 <= candidates[i] <= 200，我就放心了。
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本题和[77.组合](https://programmercarl.com/0077.组合.html)，[216.组合总和III](https://programmercarl.com/0216.组合总和III.html)的区别是：本题没有数量要求，可以无限重复，但是有总和的限制，所以间接的也是有个数的限制。
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本题搜索的过程抽象成树形结构如下：
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注意图中叶子节点的返回条件，因为本题没有组合数量要求，仅仅是总和的限制，所以递归没有层数的限制，只要选取的元素总和超过target，就返回！
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而在[77.组合](https://programmercarl.com/0077.组合.html)和[216.组合总和III](https://programmercarl.com/0216.组合总和III.html) 中都可以知道要递归K层，因为要取k个元素的组合。
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## 回溯三部曲
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* 递归函数参数
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这里依然是定义两个全局变量，二维数组result存放结果集，数组path存放符合条件的结果。（这两个变量可以作为函数参数传入）
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首先是题目中给出的参数，集合candidates, 和目标值target。
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此外我还定义了int型的sum变量来统计单一结果path里的总和，其实这个sum也可以不用，用target做相应的减法就可以了，最后如何target==0就说明找到符合的结果了，但为了代码逻辑清晰，我依然用了sum。
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**本题还需要startIndex来控制for循环的起始位置，对于组合问题，什么时候需要startIndex呢？**
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我举过例子，如果是一个集合来求组合的话，就需要startIndex，例如：[77.组合](https://programmercarl.com/0077.组合.html)，[216.组合总和III](https://programmercarl.com/0216.组合总和III.html)。
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如果是多个集合取组合，各个集合之间相互不影响，那么就不用startIndex，例如：[17.电话号码的字母组合](https://programmercarl.com/0017.电话号码的字母组合.html)
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**注意以上我只是说求组合的情况，如果是排列问题，又是另一套分析的套路，后面我再讲解排列的时候就重点介绍**。
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代码如下：
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```CPP
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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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```
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* 递归终止条件
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在如下树形结构中：
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从叶子节点可以清晰看到，终止只有两种情况，sum大于target和sum等于target。
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sum等于target的时候，需要收集结果，代码如下：
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```CPP
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if (sum > target) {
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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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* 单层搜索的逻辑
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单层for循环依然是从startIndex开始，搜索candidates集合。
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**注意本题和[77.组合](https://programmercarl.com/0077.组合.html)、[216.组合总和III](https://programmercarl.com/0216.组合总和III.html)的一个区别是：本题元素为可重复选取的**。
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如何重复选取呢，看代码，注释部分：
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```CPP
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for (int i = startIndex; i < candidates.size(); i++) {
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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); // 关键点:不用i+1了，表示可以重复读取当前的数
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sum -= candidates[i]; // 回溯
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path.pop_back(); // 回溯
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}
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```
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按照[关于回溯算法，你该了解这些！](https://programmercarl.com/回溯算法理论基础.html)中给出的模板，不难写出如下C++完整代码：
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```CPP
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// 版本一
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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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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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for (int i = startIndex; i < candidates.size(); i++) {
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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); // 不用i+1了，表示可以重复读取当前的数
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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>> combinationSum(vector<int>& candidates, int target) {
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result.clear();
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path.clear();
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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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## 剪枝优化
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在这个树形结构中：
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以及上面的版本一的代码大家可以看到，对于sum已经大于target的情况，其实是依然进入了下一层递归，只是下一层递归结束判断的时候，会判断sum > target的话就返回。
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其实如果已经知道下一层的sum会大于target，就没有必要进入下一层递归了。
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那么可以在for循环的搜索范围上做做文章了。
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**对总集合排序之后，如果下一层的sum（就是本层的 sum + candidates[i]）已经大于target，就可以结束本轮for循环的遍历**。
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如图：
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for循环剪枝代码如下：
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```
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for (int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++)
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```
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整体代码如下：（注意注释的部分）
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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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// 如果 sum + candidates[i] > target 就终止遍历
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for (int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++) {
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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);
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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>> combinationSum(vector<int>& candidates, int target) {
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result.clear();
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path.clear();
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sort(candidates.begin(), candidates.end()); // 需要排序
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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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* 时间复杂度: O(n * 2^n)，注意这只是复杂度的上界，因为剪枝的存在，真实的时间复杂度远小于此
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* 空间复杂度: O(target)
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# 总结
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本题和我们之前讲过的[77.组合](https://programmercarl.com/0077.组合.html)、[216.组合总和III](https://programmercarl.com/0216.组合总和III.html)有两点不同：
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* 组合没有数量要求
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* 元素可无限重复选取
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针对这两个问题，我都做了详细的分析。
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并且给出了对于组合问题，什么时候用startIndex，什么时候不用，并用[17.电话号码的字母组合](https://programmercarl.com/0017.电话号码的字母组合.html)做了对比。
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最后还给出了本题的剪枝优化，这个优化如果是初学者的话并不容易想到。
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**在求和问题中，排序之后加剪枝是常见的套路！**
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可以看出我写的文章都会大量引用之前的文章，就是要不断作对比，分析其差异，然后给出代码解决的方法，这样才能彻底理解题目的本质与难点。
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# 其他语言版本
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## Java
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```Java
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// 剪枝优化
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class Solution {
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public List<List<Integer>> combinationSum(int[] candidates, int target) {
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List<List<Integer>> res = new ArrayList<>();
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Arrays.sort(candidates); // 先进行排序
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backtracking(res, new ArrayList<>(), candidates, target, 0, 0);
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return res;
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}
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public void backtracking(List<List<Integer>> res, List<Integer> path, int[] candidates, int target, int sum, int idx) {
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// 找到了数字和为 target 的组合
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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 = idx; i < candidates.length; i++) {
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// 如果 sum + candidates[i] > target 就终止遍历
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if (sum + candidates[i] > target) break;
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path.add(candidates[i]);
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backtracking(res, path, candidates, target, sum + candidates[i], i);
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path.remove(path.size() - 1); // 回溯，移除路径 path 最后一个元素
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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（版本一）
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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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return
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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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total += candidates[i]
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path.append(candidates[i])
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self.backtracking(candidates, target, total, i, path, result) # 不用i+1了，表示可以重复读取当前的数
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total -= candidates[i]
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path.pop()
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def combinationSum(self, candidates, target):
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result = []
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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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回溯剪枝（版本一）
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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 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, path, result)
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total -= candidates[i]
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path.pop()
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def combinationSum(self, candidates, target):
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result = []
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candidates.sort() # 需要排序
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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（版本二）
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```python
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class Solution:
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def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
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result =[]
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self.backtracking(candidates, target, 0, [], result)
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return result
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def backtracking(self, candidates, target, startIndex, path, result):
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if target == 0:
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result.append(path[:])
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return
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if target < 0:
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return
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for i in range(startIndex, len(candidates)):
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path.append(candidates[i])
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self.backtracking(candidates, target - candidates[i], i, path, result)
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path.pop()
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```
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回溯剪枝（版本二）
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```python
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class Solution:
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def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
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result =[]
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candidates.sort()
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self.backtracking(candidates, target, 0, [], result)
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return result
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def backtracking(self, candidates, target, startIndex, path, result):
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if target == 0:
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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 target - candidates[i] < 0:
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break
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path.append(candidates[i])
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self.backtracking(candidates, target - candidates[i], i, path, result)
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path.pop()
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```
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## Go
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主要在于递归中传递下一个数字
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```go
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var (
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res [][]int
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path []int
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)
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func combinationSum(candidates []int, target int) [][]int {
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res, path = make([][]int, 0), make([]int, 0, len(candidates))
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sort.Ints(candidates) // 排序，为剪枝做准备
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dfs(candidates, 0, target)
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return res
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}
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func dfs(candidates []int, start int, target int) {
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if target == 0 { // target 不断减小，如果为0说明达到了目标值
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tmp := make([]int, len(path))
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copy(tmp, path)
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res = append(res, tmp)
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return
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}
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for i := start; i < len(candidates); i++ {
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if candidates[i] > target { // 剪枝，提前返回
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break
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}
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path = append(path, candidates[i])
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dfs(candidates, i, target - candidates[i])
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path = path[:len(path) - 1]
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}
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}
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```
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## JavaScript
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```js
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var combinationSum = function(candidates, target) {
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const res = [], path = [];
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candidates.sort((a,b)=>a-b); // 排序
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backtracking(0, 0);
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return res;
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function backtracking(j, sum) {
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if (sum === target) {
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res.push(Array.from(path));
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return;
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}
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for(let i = j; i < candidates.length; i++ ) {
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const n = candidates[i];
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if(n > target - sum) break;
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path.push(n);
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sum += n;
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backtracking(i, sum);
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path.pop();
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sum -= n;
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}
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}
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};
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```
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## TypeScript
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```typescript
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function combinationSum(candidates: number[], target: number): number[][] {
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const resArr: number[][] = [];
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function backTracking(
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candidates: number[], target: number,
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startIndex: number, route: number[], curSum: number
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): void {
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if (curSum > target) return;
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if (curSum === target) {
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resArr.push(route.slice());
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return
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}
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for (let i = startIndex, length = candidates.length; i < length; i++) {
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let tempVal: number = candidates[i];
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route.push(tempVal);
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backTracking(candidates, target, i, route, curSum + tempVal);
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route.pop();
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}
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}
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backTracking(candidates, target, 0, [], 0);
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return resArr;
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};
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```
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## Rust
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```Rust
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impl Solution {
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pub fn backtracking(result: &mut Vec<Vec<i32>>, path: &mut Vec<i32>, candidates: &Vec<i32>, target: i32, mut sum: i32, start_index: usize) {
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if sum == target {
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result.push(path.to_vec());
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return;
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}
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for i in start_index..candidates.len() {
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if sum + candidates[i] <= target {
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sum += candidates[i];
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path.push(candidates[i]);
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Self::backtracking(result, path, candidates, target, sum, i);
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sum -= candidates[i];
|
||
path.pop();
|
||
}
|
||
}
|
||
}
|
||
|
||
pub fn combination_sum(candidates: Vec<i32>, target: i32) -> Vec<Vec<i32>> {
|
||
let mut result: Vec<Vec<i32>> = Vec::new();
|
||
let mut path: Vec<i32> = Vec::new();
|
||
Self::backtracking(&mut result, &mut path, &candidates, target, 0, 0);
|
||
result
|
||
}
|
||
}
|
||
```
|
||
|
||
## C
|
||
|
||
```c
|
||
int* path;
|
||
int pathTop;
|
||
int** ans;
|
||
int ansTop;
|
||
//记录每一个和等于target的path数组长度
|
||
int* length;
|
||
|
||
void backTracking(int target, int index, int* candidates, int candidatesSize, int sum) {
|
||
//若sum>=target就应该终止遍历
|
||
if(sum >= target) {
|
||
//若sum等于target，将当前的组合放入ans数组中
|
||
if(sum == target) {
|
||
int* tempPath = (int*)malloc(sizeof(int) * pathTop);
|
||
int j;
|
||
for(j = 0; j < pathTop; j++) {
|
||
tempPath[j] = path[j];
|
||
}
|
||
ans[ansTop] = tempPath;
|
||
length[ansTop++] = pathTop;
|
||
}
|
||
return ;
|
||
}
|
||
|
||
int i;
|
||
for(i = index; i < candidatesSize; i++) {
|
||
//将当前数字大小加入sum
|
||
sum+=candidates[i];
|
||
path[pathTop++] = candidates[i];
|
||
backTracking(target, i, candidates, candidatesSize, sum);
|
||
sum-=candidates[i];
|
||
pathTop--;
|
||
}
|
||
}
|
||
|
||
int** combinationSum(int* candidates, int candidatesSize, int target, int* returnSize, int** returnColumnSizes){
|
||
//初始化变量
|
||
path = (int*)malloc(sizeof(int) * 50);
|
||
ans = (int**)malloc(sizeof(int*) * 200);
|
||
length = (int*)malloc(sizeof(int) * 200);
|
||
ansTop = pathTop = 0;
|
||
backTracking(target, 0, candidates, candidatesSize, 0);
|
||
|
||
//设置返回的数组大小
|
||
*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 combinationSum(_ candidates: [Int], _ target: Int) -> [[Int]] {
|
||
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 {
|
||
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) // i不用+1以重复访问
|
||
path.removeLast() // 回溯
|
||
}
|
||
}
|
||
backtracking(sum: 0, startIndex: 0)
|
||
return result
|
||
}
|
||
```
|
||
|
||
## Scala
|
||
|
||
```scala
|
||
object Solution {
|
||
import scala.collection.mutable
|
||
def combinationSum(candidates: Array[Int], target: Int): List[List[Int]] = {
|
||
var result = mutable.ListBuffer[List[Int]]()
|
||
var path = mutable.ListBuffer[Int]()
|
||
|
||
def backtracking(sum: Int, index: Int): Unit = {
|
||
if (sum == target) {
|
||
result.append(path.toList) // 如果正好等于target，就添加到结果集
|
||
return
|
||
}
|
||
// 应该是从当前索引开始的，而不是从0
|
||
// 剪枝优化：添加循环守卫，当sum + c(i) <= target的时候才循环，才可以进入下一次递归
|
||
for (i <- index until candidates.size if sum + candidates(i) <= target) {
|
||
path.append(candidates(i))
|
||
backtracking(sum + candidates(i), i)
|
||
path = path.take(path.size - 1)
|
||
}
|
||
}
|
||
|
||
backtracking(0, 0)
|
||
result.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>
|