Leetcode--Combination Sum II

Problem Description:

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] 

分析：这题和Combination Sum题目的差别在于每一个元素仅仅能使用一次，因此在深度搜索结果时要注意去掉反复的结果，考虑一下事实上仅仅须要简单的修改两个地方就可以，第一个是在循环时增加推断是否与前一个元素同样的情况，同样则不再考虑，第二个是递归时每一个元素仅仅能使用一次。

代码例如以下：

class Solution {
public:

    void dfs(vector<int> &candidates,int target,int start,vector<vector<int> > &results,vector<int> &subset)
    {
        if(target==0)
        {
            results.push_back(subset);
            return;
        }
        if(target<0||start>=candidates.size()) 
            return;
        for(size_t i=start;i!=candidates.size();++i)
        {
            if(i>start && candidates[i]==candidates[i-1]) continue;
            subset.push_back(candidates[i]);
            dfs(candidates,target-candidates[i],i+1,results,subset);
            subset.pop_back();
        }
        
    }

    vector<vector<int> > combinationSum2(vector<int> &num, int target) {
        sort(num.begin(),num.end());
        vector<vector<int> > results;
        vector<int> subset;
        dfs(num,target,0,results,subset);
        return results;
    }
};