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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]
Analyse: First sort the vector, then recursively invoke the dfs function.
Runtime: 16ms.
1 class Solution { 2 public: 3 vector<vector<int>> combinationSum(vector<int>& candidates, int target) { 4 sort(candidates.begin(), candidates.end()); 5 vector<vector<int> >result; 6 vector<int> temp; 7 8 dfs(temp, candidates, target, 0, result); 9 return result; 10 } 11 12 void dfs(vector<int> &temp, vector<int> &candidates, int diff, int start, vector<vector<int> > &result){ 13 if(diff == 0){ 14 result.push_back(temp); 15 return ; 16 } 17 18 for(int i = start; i < candidates.size(); i++){ 19 if(diff < candidates[i]) return; 20 21 temp.push_back(candidates[i]); 22 dfs(temp, candidates, diff - candidates[i], i, result); 23 temp.pop_back(); 24 } 25 } 26 };
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原文地址:http://www.cnblogs.com/amazingzoe/p/4678438.html