LeetCode - 40. Combination Sum II40. Combination Sum II Problem's Link

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Mean: 

给你一个待选集合s和一个数n，选出所有相加之和为n的组合.（每个元素只能选一次）

analyse:

递归求解.

在递归进入下一层时加判断，相邻两个相等则跳过一次，否则如果当前值对应后面组合有答案，将会算进去两次.

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>> combinationSum2(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)

           if(i==begin || candidates[i]!=candidates[i-1])

           {

               combination.push_back(candidates[i]);

               combinationSum(candidates,res,combination,target-candidates[i],i+1);

               combination.pop_back();

           }

       return;

};

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.combinationSum2(ve,target);

       for(auto p1:ans)

           for(auto p2:p1)

               cout<<p2<<" ";

           cout<<endl;

   return 0;

}

/*