39. Combination Sum

description:

给定target， 求给定数列中找到几个数（其中的数可以重复使用，且一组数有几个也不做限制）的和为target
Note:

• https://www.cnblogs.com/grandyang/p/4606334.html

Example:

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

answer:

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        vector<int> out;
        combinationSumDFS(candidates, target, 0, out, res);
        return res; 
        // return combinationSumDFS(candidates, target, 0, out, res); 这是我之前的写法，注意错误！！！
    }
    void combinationSumDFS(vector<int>& candidates, int target, int start, vector<int>& out, vector<vector<int>>& res){
        if (target < 0) return;
        if (target == 0) {res.push_back(out); return;}
        for (int i = start; i < candidates.size(); ++i) {
            out.push_back(candidates[i]);
            combinationSumDFS(candidates, target - candidates[i], i, out, res);
            out.pop_back();
        }
    }
};

relative point get√：

像这种结果要求返回所有符合要求解的题十有八九都是要利用到递归，而且解题的思路都大同小异，相类似的题目有 Path Sum II，Subsets II，Permutations，Permutations II，Combinations 等等，如果仔细研究这些题目发现都是一个套路，都是需要另写一个递归函数

hint :