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Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:
For example, given candidate set 10,1,2,7,6,1,5 and target 8,
A solution set is: [1, 7] [1, 2, 5] [2, 6] [1, 1, 6]
class Solution { public: void dfs(const vector<int>& candidates,int target,vector<vector<int> >& res,vector<int>& dummy,int index){ if(target<0){ return; } else if(target==0){ for(int i=0;i<res.size();i++) if(res[i]==dummy) return; res.push_back(dummy); return; } for(int i=index;i<candidates.size();i++){ dummy.push_back(candidates[i]); dfs(candidates,target-candidates[i],res,dummy,i+1); dummy.pop_back();//典型的回溯思想 } } vector<vector<int> > combinationSum2(vector<int>& candidates, int target) { sort(candidates.begin(),candidates.end()); vector<vector<int> >res; vector<int> dummy; dfs(candidates,target,res,dummy,0); return res; } };
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原文地址：http://www.cnblogs.com/wqkant/p/5357419.html
