Combination Sum II

题目：
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:
All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.
For example, given candidate set 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
思路：
1、和上面的那个题一样，只是在递归的时候变成了

comb(candidates,i+1,sum+candidates[i],target,res,path);  

代码：

class Solution {public:    void comb(vector<int> candidates, int index, int sum, int target, vector<vector<int>> &res, vector<int> &path)      {          if(sum>target)return;          if(sum==target){res.push_back(path);return;}          for(int i= index; i<candidates.size();i++)          {              path.push_back(candidates[i]);              comb(candidates,i+1,sum+candidates[i],target,res,path);              path.pop_back();            while(i<candidates.size() && candidates[i]==candidates[i+1]) i++;        }      }      vector<vector<int> > combinationSum2(vector<int> &candidates, int target) {          // Note: The Solution object is instantiated only once.          sort(candidates.begin(),candidates.end());          vector<vector<int>> res;          vector<int> path;          comb(candidates,0,0,target,res,path);          return res;      }  };