[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：

 1 class Solution {
 2 public:
 3     vector<vector<int> > combinationSum(vector<int> &candidates, int target) 
 4     {
 5         vector<vector<int>> res;
 6         vector<int> midVal;
 7         sort(candidates.begin(),candidates.end());
 8         DFS(res,midVal,candidates,target,0);
 9         return res;    
10     }
11 
12     void DFS(vector<vector<int>> &res,vector<int> &midVal,vector<int> &candidates,int target,int start)
13     {
14         if(target<0)
15             return;
16         else if(target==0)
17             res.push_back(midVal);
18         else
19         {
20             for(int i=start;i<candidates.size();i++)
21             {
22                 
23                 midVal.push_back(candidates[i]);
24                 DFS(res,midVal,candidates,target-candidates[i],i);  // i
25                 midVal.pop_back();
26             }
27         }
28     }
29 };