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] 

思路：这道题是Combination Sum的延伸，不能有重复的结果集。只需在I上修改，先统计出C中每个数字的出现次数，

在组合时如果用到该数字则将数字的次数减一。

class Solution {private:    map<int,int> numFreq;    vector<vector<int> > res;    vector<int> perRes;public:    void combinationSum2Helper(int target, int pos, int left)    {        if (target == 0)        {            res.push_back(perRes);        }         else         {            map<int,int>::iterator it;            for(it=numFreq.begin(); it!=numFreq.end(); ++it)            {                if(it->first>=left && it->second>0 && target-it->first>=0)                {                    it->second--;                    perRes.resize(pos+1);                    perRes[pos] = it->first;                    combinationSum2Helper(target-it->first,pos+1,it->first);                    it->second++;                }                }            }           }        vector<vector<int> > combinationSum2(vector<int> &num, int target) {        numFreq.clear();        res.clear();        perRes.clear();        if (num.empty())        {            return res;        }            int len = num.size();        int i;        for(i=0; i<len; ++i)        {            if (numFreq.find(num[i]) != numFreq.end())            {                numFreq[num[i]]++;            }                else             {                numFreq.insert(pair<int,int>(num[i],1));            }            }        combinationSum2Helper(target, 0, 0);         return res;       }};