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题目：
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. （Medium）
Note:
For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:
[ [7], [2, 2, 3] ]
分析：
先排个序，用DFS搜索，每个数可选可不选，然后在start > end或者candidates[start] > target后就return。
恰好candidates[start] = target满足时添加到结果中。
代码：
1 class Solution { 2 private: 3 vector<vector<int>>result; 4 void dfs(int start, int end, const vector<int>& candidates, int target, vector<int>& internal) { 5 if (start > end) { 6 return; 7 } 8 if (candidates[start] == target) { 9 internal.push_back(candidates[start]); 10 result.push_back(internal); 11 internal.pop_back(); 12 return; 13 } 14 if (candidates[start] > target) { 15 return; 16 } 17 dfs(start + 1, end, candidates, target, internal); 18 internal.push_back(candidates[start]); 19 dfs(start, end, candidates, target - candidates[start], internal); 20 internal.pop_back(); 21 } 22 public: 23 vector<vector<int>> combinationSum(vector<int>& candidates, int target) { 24 sort(candidates.begin(), candidates.end()); 25 int end = candidates.size() - 1; 26 vector<int> internal; 27 dfs(0, end, candidates, target, internal); 28 return result; 29 } 30 };
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原文地址：http://www.cnblogs.com/wangxiaobao/p/5811750.html
