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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.
Note:
For example, given candidate set 2,3,6,7 and target 7,
A solution set is: [7] [2, 2, 3]
代码：
class Solution { public: vector<vector<int>> combinationSum(vector<int>& candidates, int target) { vector<vector<int> > ret; vector<int> tmp; int sum = 0; std::sort(candidates.begin(), candidates.end()); Solution::dfs(ret, tmp, sum, candidates, 0, candidates.size()-1, target); return ret; } static void dfs( vector<vector<int> >& ret, vector<int>& tmp, int &sum, vector<int>& candidates, int begin, int end, int target ) { if ( sum>target ) return; if ( sum==target ) { ret.push_back(tmp); return; } for ( int i=begin; i<=end; ++i ) { if ( sum+candidates[i]<=target ) { sum += candidates[i]; tmp.push_back(candidates[i]); Solution::dfs(ret, tmp, sum, candidates, i, end, target); sum -= candidates[i]; tmp.pop_back(); } } } };
tips：
采用深搜模板：
1. 终止条件sum>target
2. 加入解集条件sum==target
3. 遍历当前层所有分支（如果满足sum+candidates[i]<target，则还可以再往上加candidates[i]；注意，这里传入下一层的begin下标为i，因为要求元素可以无限多重复）
4. 由于传入下一层始终满足begin<=end，因此不要在终止条件中加入（begin>end）
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4535900.html
