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]

  题意：从数组中找到和为目标的组合，每个数允许重复。

  思路：排序后，利用回溯去做，为了让每个数都能重复，所以枚举的时候可以在从这个数的下标开始一次

  class Solution {
  public:
      void dfs(vector
      
        candidates, int index, int sum, int target, vector
       
        > &res, vector
        
          &path) { if (sum > target) return; if (sum == target) { res.push_back(path); return; } for (int i = index; i < candidates.size(); i++) { path.push_back(candidates[i]); dfs(candidates, i, sum+candidates[i], target, res, path); path.pop_back(); } } vector
         
           > combinationSum(vector
          
            &candidates, int target) { sort(candidates.begin(), candidates.end()); vector
           
             > res; vector
            
              path; dfs(candidates, 0, 0, target, res, path); return res; } };