Combination Sum II 题解

题目来源：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]

解题思路：

跟 combination-sum 一样, DFS，在它的基础上，将用过的跳过即可。重复的数字，后面的要跳过，不然结果有重复的。

    void dfs(vector<vector<int> > &result, vector<int> &path, const vector<int> &num, int sum, int target, int start)
    {
        if(sum > target) return;
        if(sum == target)
        {
            result.push_back(path);
            return;
        }
        for(int i = start ; i < num.size(); i++)
        {
            if(sum < target)
            {
                sum += num[i];
                path.push_back(num[i]);
                dfs(result, path, num, sum, target, i+1);
                path.pop_back();
                sum -= num[i];
            }
            while(i+1 < num.size() && num[i] == num[i+1]) i++; ////ignore the next same one,so that remove the duplicated result
        }
    }

    vector<vector<int> > combinationSum2(vector<int> &num, int target) 
    {
         vector<vector<int> > result;
        if(num.size() == 0) return result;
        std::sort(num.begin(), num.end());
        vector<int> path;
        dfs(result, path, num, 0, target, 0);
        return move(result);
    }