这题居然没有记录，写一下吧：
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]
1 // 17:10 2 class Solution { 3 private: 4 vector<vector<int> > res; 5 public: 6 vector<vector<int> > combinationSum(vector<int> &candidates, int target) { 7 res.clear(); 8 sort(candidates.begin(), candidates.end()); 9 vector<int> path; 10 dfs(candidates, candidates.size() - 1, target, path); 11 return res; 12 } 13 14 void dfs(vector<int>& candidates, int pos, int target, vector<int>& path) { 15 if (target == 0) { 16 res.push_back(path); 17 reverse(res.back().begin(), res.back().end()); 18 return; 19 } 20 if (target < 0 || pos < 0) { 21 return; 22 } 23 24 int curv = candidates[pos]; 25 26 // don‘t use current value 27 dfs(candidates, pos - 1, target, path); 28 int cnt = 0; 29 while (target > 0) { 30 path.push_back(curv); 31 target -= curv; 32 dfs(candidates, pos - 1, target, path); 33 cnt++; 34 } 35 36 path.resize(path.size() - cnt); 37 } 38 };
原文：http://www.cnblogs.com/lailailai/p/4456254.html