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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] 

给定一组候选解集合，以及一个目标值，给出所有能够累加得到该目标值的组合，候选解元素可以重复使用任意次。

题目要求解的形式要按非降序，那就开始之前将候选序列排个序。

Solution {public:    /*      cur_sum: 当前的累加和      icur   : 当前候选数的下标      target ：目标值      cur_re : 当前的潜在解      re     : 最终的结果集      can    : 候选数集    */    void dfs(int cur_sum, int icur, int target, vector
   
    & cur_re, vector
    
     
       > &re, vector
      
        can)    {        if(cur_sum > target || icur >= can.size()) return;        if(cur_sum == target){            //cur_re.push_back(cur_sum);            re.push_back(cur_re);            return;        }        //(target - cur_sum >= can[i]类似剪枝，加上该判定与        //不加，时间差4-5倍。        for(int i = icur; i < can.size() && (target - cur_sum >= can[i]); ++i){            cur_sum += can[i];            cur_re.push_back(can[i]);            dfs(cur_sum, i, target, cur_re, re, can);            //go back            cur_re.pop_back();            cur_sum -= can[i];        }    }    vector
       
        
          > combinationSum(vector
         
           &candidates, int target) { vector
          
           
             > re; vector
            
              cur_re; if(candidates.size() == 0) return re; sort(candidates.begin(), candidates.end()); if(candidates[0] > target) return re; dfs(0, 0, target, cur_re, re, candidates); return re; }};
            
           
          
         
        
       
      
     
    
   

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