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]

Analyse: First sort the vector, then recursively invoke the dfs function.

Runtime: 16ms.

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