DFS（全排列）--求子數組和為sum的所有組合方案（不包含重複元素）

Given a collection of candidate numbers (​

​candidates​

​​) and a target number (​

​target​

​​), find all unique combinations in ​

​candidates​

​​ where the candidate numbers sums to ​

​target​

​.

Each number in ​

​candidates​

​ may only be used once in the combination.

Note:

• All numbers (including​

  ​target​

  ​) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = ​

​[10,1,2,7,6,1,5]​

​​, target = ​

​8​

​,

A solution set is:

[

[1, 7],

[1, 2, 5],

[2, 6],

[1, 1, 6]

]

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]      

class Solution{
private:
    vector<vector<int> >Result;
public:
    void DFS(vector<int> candidates,int Index,vector<int>& Temp,int target)
    {
        if(target < 0){return;}
        if(target == 0){
            Result.push_back(Temp);return;
        }
        for(int i = Index;i < candidates.size();i ++){
            if(i != Index && candidates[i] == candidates[i - 1]){continue;}
            Temp.push_back(candidates[i]);
            DFS(candidates,i + 1,Temp,target - candidates[i]);
            Temp.pop_back();
        }
    }
    vector<vector<int> > combinationSum2(vector<int>& candidates,int target){
        if(candidates.size() == 0){return Result;}
        sort(candidates.begin(),candidates.end());
        vector<int> Temp;
        DFS(candidates,0,Temp,target);
        return Result;
    }

};