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:
For example, given candidate set 2,3,6,7 and target 7,
A solution set is: [7] [2, 2, 3]
分析：这道题类似combinaions和subsets，不同的是，在本题的可行解中可以存在不限数目的相同的数，同样我们用改良的dfs方法求解。因为此题要求每个combination中元素必须是非递减排列的，所以我们先将元素按升序排序。此外由于元素全为正数，当当前path的sum大于target时，即可停止该path的搜素。代码如下：
1 class Solution { 2 public: 3 vector<vector<int> > combinationSum(vector<int> &candidates, int target) { 4 vector<vector<int>> res; 5 vector<int> path; 6 sort(candidates.begin(),candidates.end()); 7 dfs(res,path,candidates,0,0,target); 8 return res; 9 } 10 void dfs(vector<vector<int>> &res, vector<int> & path, vector<int> & c, int start, int sum, int target){ 11 for(int i = start; i < c.size(); i++){ 12 path.push_back(c[i]); 13 if(sum + c[i] == target)res.push_back(path); 14 else if(sum + c[i] < target){ 15 dfs(res,path,c,i,sum+c[i],target); 16 } 17 path.pop_back(); 18 } 19 } 20 };
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原文：http://www.cnblogs.com/Kai-Xing/p/3916538.html