Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

It is guaranteed that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

Example 1:

Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.

Example 2:

Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3:****

Input: candidates = [2], target = 1
Output: []

Example 4:

Input: candidates = [1], target = 1
Output: [[1]]

Example 5:

Input: candidates = [1], target = 2
Output: [[1,1]]

Constraints:

1 <= candidates.length <= 30
1 <= candidates[i] <= 200
All elements of candidates are distinct.
1 <= target <= 500

实现思路：
题意给定一个序列，然后要求按照元素增序要求，求出所有元素组成和为目标值的结果序列，初始元素可以重复选择，这道题依旧采用PAT中这道https://www.cnblogs.com/coderJ-one/p/14409000.html题目的思路，采用深搜+剪枝的方式去求解所有的可能即可。

AC代码：

class Solution {
		vector<int> temp;
		vector<vector<int>> ans;
		void dfs(int idx,int sum,int target,int k,vector<int> &sq) {
			if(sum==target) {
				ans.push_back(temp);
				return;
			}
			if(sum>target||idx>=k) return;//剪枝 当数的总和已经大于目标数或者当数组索引下标超出范围的时候就退出
			if(sq[idx]+sum<=target) {
				temp.push_back(sq[idx]);
				dfs(idx,sum+sq[idx],target,k,sq);
				temp.pop_back();
				dfs(idx+1,sum,target,k,sq);
			}
		}

	public:
		vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
			sort(candidates.begin(),candidates.end());
			dfs(0,0,target,candidates.size(),candidates);
			return ans;
		}
};