LeetCode

39. Combination Sum 

Problem's Link

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Mean: 

给你一个待选集合s和一个数n,让你找出集合s中相加之和为n的所有组合.（每个数可选多次）

analyse:

作此题需对递归有一定的理解.

Time complexity: O(N)

 

view code

/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
*       Author: crazyacking
*       Date  : 2016-03-05-18.56
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);

class Solution
{
public:
   vector<vector<int>> combinationSum(vector<int>& candidates, int target)
   {
       sort(candidates.begin(),candidates.end());
       vector<vector<int>> res;
       vector<int> combination;
       combinationSum(candidates,res,combination,target,0);
       return res;
   }
private:
   void combinationSum(vector<int>& candidates,vector<vector<int>> &res,vector<int>& combination,int target,int begin)
   {
       if(!target)
       {
           res.push_back(combination);
           return;
       }
       for(int i=begin;target>=candidates[i] && i<candidates.size() ;++i)
       {
           combination.push_back(candidates[i]);
           combinationSum(candidates,res,combination,target-candidates[i],i);
           combination.pop_back();
       }
   }
};

int main()
{
   freopen("H:\Code_Fantasy\in.txt","r",stdin);
   int n,target;
   while(cin>>n>>target)
   {
       cout<<n<<" "<<target<<endl;
       vector<int> ve;
       for(int i=0; i<n; ++i)
       {
           int tmp;
           cin>>tmp;
           ve.push_back(tmp);
       }
       Solution solution;
       vector<vector<int>> ans=solution.combinationSum(ve,target);
       for(auto p1:ans)
       {
           for(auto p2:p1)
           {
               cout<<p2<<" ";
           }
           cout<<endl;
       }
   }
   return 0;
}
/*

*/