标签:des style blog color os io strong for ar
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]
方法:用queue实现bfs
class Solution { public: vector<vector<int> > combinationSum(vector<int> &candidates, int target) { vector<vector<int> > result; if(candidates.size()==0){ vector<int> tmp; result.push_back(tmp); return result; } sort(candidates.begin(),candidates.end()); bfs(result,candidates,target); return result; }//end func private: void bfs(vector<vector<int> > &result,vector<int> &candidates,int target){ queue<vector<int> > q; int len = candidates.size(); for(int i = 0;i<len;i++){ int sum = 0; int value = candidates[i]; vector<int> tmp; while(true){ sum += value; tmp.push_back(value); if(sum<target){ q.push(tmp); }else if(sum == target){ if(find(result.begin(),result.end(),tmp)==result.end()) result.push_back(tmp); } else break; } while(!q.empty()){ tmp = q.front(); q.pop(); sum = 0; for(int k=0;k<tmp.size();k++){ sum += tmp[k]; } int sum0 = sum; vector<int> tmp0(tmp); for(int j=i+1;j<len;j++){ value = candidates[j]; while(true){ sum += value; tmp.push_back(value); if(sum<target){ q.push(tmp); }else if(sum == target){ sort(tmp.begin(),tmp.end()); if(find(result.begin(),result.end(),tmp)==result.end()) result.push_back(tmp); } else break; } sum = sum0; tmp = tmp0; } } }//end for }//end func };
[LeetCode] Combination Sum (bfs)
标签:des style blog color os io strong for ar
原文地址:http://www.cnblogs.com/Xylophone/p/3933636.html