【算法学习记录-散列】【PAT A1041】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]. 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 (≤) 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

思路：

0、《算法笔记》给出的参考答案是使用两个数组，作者使用的是二维数组，相比更繁杂

 

1、整体思路

建立arr[10000][2]，其中第一列存放下标数值出现的次数，第二列存放下标数值出现的顺序（参考答案是用另外一个数字存放输入内容，从而得到顺序）

首先，for循环输入所有数据以及出现顺序，注意需要用if判断“顺序”属性是否是第一次输入

而后，for循环寻找只出现一次且是第一个被输入的数据

 

2、题解代码

#include<cstdio>

int main() {
    
    int arr[10000][2];
    for (int i = 0; i < 10000; i++) {
        arr[i][0] = 0;
        arr[i][1] = 100010;
    }
    
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        int num;
        scanf("%d", &num);
        arr[num][0]++;
        if (arr[num][1] == 100010) {
            arr[num][1] = i;
        }
    }
    
    int k = 100010, win = 0;//为什么是100010？首先因为一共最多有100000个数据，然后这是用arr[i][1] < k的逻辑来寻找第一个出现的数据，所以不能赋值0
    for (int i = 0; i < 10000; i++) {
        if (arr[i][0] == 1 && arr[i][1] < k) {
            win = i;
            k = arr[i][1];
        }
    }
    
    if (win != 0) {
        printf("%d", win);
    } else {
        printf("None");
    }
    
    return 0;
}