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]
vector<vector<int> > combinationSum(vector<int> &candidates, int target) { vector<vector<int> >res; vector<int>temp; sort(candidates.begin(),candidates.end()); dfs(candidates,target,0,temp,res); return res;}private:void dfs(vector<int>& candidates, int gap, int idx, vector<int>& temp, vector<vector<int> >& res){ if(gap == 0){ res.push_back(temp); return; } for(int i=idx; i<candidates.size(); i++){//皇帝轮流 if(candidates[i] > gap){//超出了 return; } temp.push_back(candidates[i]);//要了 dfs(candidates,gap-candidates[i],i,temp,res);//至少要从本身开始，不然会重复,注意是i temp.pop_back(); }}
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