[Leetcode] combination sum 组合之和

Given a set of candidate numbers ( C ) and a target number ( T ), find all unique combinations in C where the candidate numbers sums to T .

The same repeated number may be chosen from C unlimited number of times.

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a 1, a 2, … , a k) must be in non-descending order. (ie, a 1 ≤ a 2 ≤ … ≤ a k).
• The solution set must not contain duplicate combinations.

For example, given candidate set2,3,6,7and target7, 
A solution set is: 
[7]
[2, 2, 3]

题意：给定数T，在数组C中找到等于T的组合。元素可以重复利用。

思路：这题的思想和combination sum ii 基本是一样的，就是要注意的是，某一个元素本身可以重复利用，这就导致了下次深搜不是从该元素的下一个元素开始，应该是从该元素本身开始。

代码如下：

class Solution {
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) 
    {
        vector<vector<int>> res;
        vector<int> midVal;
        sort(candidates.begin(),candidates.end());
        DFS(res,midVal,candidates,target,0);
        return res;    
    }

    void DFS(vector<vector<int>> &res,vector<int> &midVal,vector<int> &candidates,int target,int start)
    {
        if(target<0)
            return;
        else if(target==0)
            res.push_back(midVal);
        else
        {
            for(int i=start;i<candidates.size();i++)
            {
                
                midVal.push_back(candidates[i]);
                DFS(res,midVal,candidates,target-candidates[i],i);  // i
                midVal.pop_back();
            }
        }
    }
};