原题链接：PAT 1041 Be Unique

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 (≤10⁵ ) 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

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题目大意：

给你若干个数，返回第一个出现的只出现过一次的数，没有这样的数就返回None

方法一：计数

思路：

用a[n]存储读入的数，同事多开一个数组m[n]用于记录a[i]出现的次数。遍历一遍存数，再遍历一遍将结果输出。

C++ 代码：

// pat 1041 
#include <iostream>
using namespace std;const int maxn = 1e5 + 10;int a[maxn], m[maxn];
int n;int main(){scanf("%d", &n);for(int i = 0; i < n; i++ ){scanf("%d", &a[i]);m[a[i]]++;}bool flag = 0;for(int i = 0; i < n; i++ ){if(m[a[i]] == 1){printf("%d", a[i]);flag = 1;break;}}if(!flag)printf("None");return 0;
}

复杂度分析：

• 时间复杂度：O(n)，遍历两遍数组
• 空间复杂度：O(n)，两个数组，数组最大长度为1e5