40. Combination Sum II

就改了上一题一个条件，不能重复使用。在回溯进入下一个的函数的时候，将pos加一。而不是上一题那样不变

code

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