【PAT甲级】Be Unique

Problem Description：

Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1,10​4​​]. The first one who bets on a unique number wins. For example, if there are 7 people betting on { 5 31 5 88 67 88 17 }, then the second one who bets on 31 wins.

Input Specification:

Each input file contains one test case. Each case contains a line which begins with a positive integer N (≤10​5​​) and then followed by N bets. The numbers are separated by a space.

Output Specification:

For each test case, print the winning number in a line. If there is no winner, print None instead.

Sample Input 1:

7 5 31 5 88 67 88 17

Sample Output 1:

31

Sample Input 2:

5 888 666 666 888 888

Sample Output 2:

None

解题思路：

题目简单来说就是一句话：“输出第一个只出现一次的数”，这肯定直接无脑map啊。然而在我对map进行无脑for-each来输出第一个只出现一次的数的时候我发现我错了，C++的map会自动根据key值来进行升序排列，它并不是像Java的LinkedHashMap一样能够按照输入的顺序来排列。这就很尴尬了，那只能够对输入的序列来进行遍历，然后输出第一个只出现一次的数。

AC代码： 

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int N;
    cin >> N;
    int a[N];
    map<int,int> m;
    for(int i = 0; i < N; i++)
    {
        cin >> a[i];
        m[a[i]]++;        
    }
    bool flag = true;
    //不能这样直接无脑for-each,因为C++的map会自动根据key值来升序排列,它不像Java的LinkedHashMap一样是能按照输入的顺序排列的
    // for(auto it : m)
    // {
    //     if(it.second == 1)
    //     {
    //         cout << it.first << endl;
    //         flag = false;
    //         break;
    //     }
    // }
    for(int i = 0; i < N; i++)
    {
        if(m[a[i]] == 1)   //输出第一个只出现过一次的数
        {
            cout << a[i] << endl;
            flag = false;
            break;
        }
    }
    if(flag)
    {
        cout << "None" << endl;
    }
    return 0;
}