LeetCode

1. Two Sum 

Problem's Link

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Mean: 

给定一个数组nums和一个数target，求:id1和id2，id1和id2为数组nums两个不同的下标，使得nums[id1]+nums[id2]=target.

注意：nums中元素可重.

analyse:

如果nums中没有重复元素，那么可以用map做.

由于有重复元素，需要用multimap.

注意：使用map时，需要用count(key)来检测是否存在key值，否则会出现错误.

Time complexity: O(N*logN)

 

view code

 

/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
*       Author: crazyacking
*       Date  : 2015-01-29-14.24
*/
#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<int> twoSum(vector<int>& nums, int target)
   {
           int cnt=0;
           vector<int> ans;
           multimap<int,int> mp;
           for(auto p:nums)
                 mp.insert(make_pair(p,cnt++));
           for(auto p:mp)
           {
                 if(mp.count(target-p.first)>0)
                 {
                       multimap<int,int>::iterator it1,it2;
                       if((p.first==target-p.first)&&(mp.count(p.first)==2))
                       {
                             it1=mp.find(p.first);
                             ans.push_back((*it1).second+1);
                             mp.erase(it1);
                             it2=mp.find(p.first);
                             ans.push_back((*it2).second+1);
                             break;
                       }
                       it1=mp.find(p.first);
                       it2=mp.find(target-p.first);
                       ans.push_back((*it1).second+1);
                       ans.push_back((*it2).second+1);
                       break;
                 }
           }
           sort(ans.begin(),ans.end());
           return ans;
   }
};

int main()
{
     int n;
     while(cin>>n)
     {
           vector<int> ve;
           for(int i=0,tmp;i<n;++i)
           {
                 cin>>tmp;
                 ve.push_back(tmp);
           }
           int target;
           cin>>target;
           Solution a;
           vector<int> ans=a.twoSum(ve,target);
           for(auto p : ans)
                 cout<<p<<endl;
     }
     return 0;
}