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:
2,3,6,7
and target
7
, [7]
[2, 2, 3]
回溯法求解。具体请参见代码。
class Solution { //C++ public: vector<int> record; vector<vector<int> >result; set<vector<int> > myset; void addSolution(){ vector<int> tmp(record.begin(),record.end()); sort(tmp.begin(),tmp.end()); if(myset.find(tmp) == myset.end()){ result.push_back(tmp); myset.insert(tmp); } } void subCombinationSum(vector<int> &cadidates, int target,int bpos){ if(target ==0 ){ addSolution(); } int size = cadidates.size(); for(int i = bpos; i < size&&cadidates[i] <= target; i++){ record.push_back(cadidates[i]); subCombinationSum(cadidates,target-cadidates[i],bpos); record.pop_back(); } } vector<vector<int> > combinationSum(vector<int> &candidates, int target) { sort(candidates.begin(),candidates.end()); subCombinationSum(candidates,target,0); return result; } };
原文地址:http://blog.csdn.net/chenlei0630/article/details/42085961