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给定一个候选人编号的集合 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。

candidates 中的每个数字在每个组合中只能使用 一次 。

注意：解集不能包含重复的组合。 

示例 1:

输入: candidates = [10,1,2,7,6,1,5], target = 8,
输出:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]

示例 2:

输入: candidates = [2,5,2,1,2], target = 5,
输出:
[
[1,2,2],
[5]
]

提示:

• 1 <= candidates.length <= 100
• 1 <= candidates[i] <= 50
• 1 <= target <= 30

class Solution {
public:vector<int> path;vector<vector<int>> res;int sum=0;void func(vector<int>& candidates, int target,int index){if(sum>target)return;if(sum==target){res.push_back(path);return;}for(int i=index;i<candidates.size();i++){if(i>index&&candidates[i]==candidates[i-1])//不是通过递归进入的循环时，i>indexcontinue;sum+=candidates[i];path.push_back(candidates[i]);func(candidates,target,i+1);sum-=candidates[i];path.pop_back();}}vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {sort(candidates.begin(),candidates.end());func(candidates,target,0);return res;}
};