[Array]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.
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]
]

方法：递归调用即可

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