【PAT甲级 - C++题解】1041 Be Unique

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专栏地址：PAT题解集合
原题地址：题目详情 - 1041 Be Unique (pintia.cn)
中文翻译：独一无二
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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 [1,10⁴]. 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

题意

找出第一个只出现一次的数字。

思路

具体思路如下：

1. 先将所有元素放入数组 a 中，并用哈希表 c 来记录每个元素出现的次数。
2. 从头往后遍历 a 中元素，输出第一个在 c 中查找数值为 1 的元素，如果不存在则输出 None 。

代码

#include
using namespace std;

const int N = 100010;
int a[N], c[N];
int n;

int main()
{
    cin >> n;

    //记录每个数出现的次数
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
        c[a[i]]++;
    }

    //找到最左边只出现一次的数字
    for (int i = 0; i < n; i++)
        if (c[a[i]] == 1)
        {
            cout << a[i] << endl;
            return 0;
        }
    puts("None");

    return 0;
}