Combination Sum II

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]

class Solution {public:    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        vector<vector<int>>res;        sort(candidates.begin(),candidates.end());        if(candidates.empty()||target<candidates.front()) //数组为空或者是target小于最小的数组元素            return res;        vector<int>slo;        compute(0,target,candidates,slo,res);        return res;    }        void compute(int startIndex,int target,vector<int>&candidates,vector<int>&slo,vector<vector<int>>&res)    {       if(target<0)           return;       if(target==0)       {           res.push_back(slo);           return;       }       for(int i=startIndex;i<candidates.size();i++)       {                      //消除在数组中重复出现的数字           if(i>startIndex&&candidates[i]==candidates[i-1]) continue;           if(target>=candidates[i])           {                slo.push_back(candidates[i]);                compute(i+1,target-candidates[i],candidates,slo,res);                slo.pop_back();                }       }    }};