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[LeetCode] Combination Sum (bfs)
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]
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 funcprivate: 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 =http://www.mamicode.com/ 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)
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