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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]
Array Backtracking
#include <iostream> #include <vector> #include <algorithm> using namespace std; class Solution { public: vector<vector<int > > ret; vector<vector<int> > combinationSum(vector<int> &candidates, int target) { vector<int > stk; ret.clear(); vector<int > tmp(candidates.begin(),candidates.end()); sort(tmp.begin(),tmp.end()); helpFun(tmp,target,0,stk); return ret; } void helpFun(vector<int> & cand,int tar, int idx,vector<int > & stk) { if(tar<0) return ; if(tar==0){ ret.push_back(stk); return ; } if(idx==cand.size()) return; stk.push_back(cand[idx]); helpFun(cand,tar-cand[idx],idx,stk); stk.pop_back(); helpFun(cand,tar,idx+1,stk); } }; int main() { vector<int > cand = {8,7,4,3}; Solution sol; vector<vector<int > > ret=sol.combinationSum(cand,11); for(int i =0;i<ret.size();i++){ for(int j=0;j<ret[i].size();j++) cout<<ret[i][j]<<" "; cout<<endl; } return 0; }
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原文地址：http://www.cnblogs.com/Azhu/p/4391404.html
