标题：
通过率：	27.7%
难度：	中等

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 (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 2,3,6,7 and target 7, 

A solution set is: 

[7] 

[2, 2, 3] 

本题依然是递归算法，算法的关键点是递归时重复元素。题目上说可以从candidate中重复一个元素，那么再递归时不用从i+1开始，还是从i开始，结果还可能是重复的情况，那么还是要res。cantain一次，具体细节看代码：

1 public class Solution { 2     public ArrayList
     
      
       > combinationSum(int[] candidates, int target) { 3         ArrayList
       
        
         > res=new ArrayList
         
          
           >(); 4 ArrayList
           
             tmp=new ArrayList
            
             (); 5 Arrays.sort(candidates); 6 dfs(candidates,target,0,res,tmp); 7 return res; 8 } 9 public void dfs(int[] candidates, int target,int start,ArrayList
             
              
               > res,ArrayList
               
                 tmp){10 if(target<0)return;11 if(target==0 && !res.contains(tmp)){12 res.add(new ArrayList
                
                 (tmp));13 return ;14 }15 for(int i=start;i