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]]

思路一：与前一题十分相似，不过存在以下几点不同。在solution set中不允许set中过的数字重复出现，以及candidate set中存在重复数字可能导致solution set中出现重复，需要进行去重

代码如下：

class Solution {public:    void combination2(vector<int>& candidates,int target,vector<int>& combination,vector<vector<int>>&res,int begin )    {        if(target == 0)        {            res.push_back(combination);            return;        }                for(int i=begin;i<candidates.size();i++)        {                        if(target - candidates[i] < 0 )                return;            if(i>begin && candidates[i] == candidates[i-1] )                continue;            combination.push_back(candidates[i]);            combination2(candidates,target-candidates[i],combination,res,i+1);            combination.pop_back();        }    }    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        sort(candidates.begin(),candidates.end());        vector<vector<int>> res;        vector<int> combination;        combination2(candidates,target,combination,res,0);        return res;    }};