[leetcode] 40. Combination Sum II 解题报告

题目链接：https://leetcode.com/problems/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：和39. Combination Sum差不多，只是这个每个数只能用一次，值得注意的是相同的数如果不做判断可能会产生相同的序列。因此在同一层的递归中，应当跳过重复的数。

代码如下：

class Solution {
public:
    void DFS(vector<int>& candidates, vector<int> vec, int target, int index, int sum)
    {
        if(sum > target)    return;
        if(sum == target)
        {
            result.push_back(vec);
            return;
        }
        for(int i = index; i< candidates.size(); i++)
        {
            vec.push_back(candidates[i]);
            DFS(candidates, vec, target, i+1, sum + candidates[i]);
            while(i < candidates.size() && candidates[i] == candidates[i+1])//去掉相同的值
                i++;
            vec.pop_back();
        }
    }

    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        if(candidates.size() == 0 || target < candidates[0])
            return result;
        vector<int> vec;
        int index = 0, sum = 0;
        DFS(candidates, vec, target, index, sum);
        return result;
    }
    
private:
    vector<vector<int>> result;
};