Combination Sum

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; } };