标签：
40. Combination Sum II
----------------------------------------------------------------------------
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;
}
/*
*/
LeetCode - 40. Combination Sum II
标签：
原文地址：http://www.cnblogs.com/crazyacking/p/5245914.html
