1056. Confusing Number 易混淆数

• 作者： 负雪明烛
• id： fuxuemingzhu
• 个人博客：https://fuxuemingzhu.cn/

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题目地址：https://leetcode-cn.com/problems/confusing-number/

题目描述

Given a number N, return true if and only if it is a confusing number, which satisfies the following condition:

We can rotate digits by 180 degrees to form new digits. When 0, 1, 6, 8, 9 are rotated 180 degrees, they become 0, 1, 9, 8, 6 respectively. When 2, 3, 4, 5 and 7 are rotated 180 degrees, they become invalid. A confusing number is a number that when rotated 180 degrees becomes a different number with each digit valid.

Example 1:

Input: 6
Output: true
Explanation:
We get 9 after rotating 6, 9 is a valid number and 9!=6.

Example 2:

Input: 89
Output: true
Explanation:
We get 68 after rotating 89, 86 is a valid number and 86!=89.

Example 3:

Input: 11
Output: false
Explanation:
We get 11 after rotating 11, 11 is a valid number but the value remains the same, thus 11 is not a confusing number.

Example 4:

Input: 25
Output: false
Explanation:
We get an invalid number after rotating 25.

Note:

1. 0 <= N <= 10^9
2. After the rotation we can ignore leading zeros, for example if after rotation we have 0008 then this number is considered as just 8.

题目大意

给定一个数字 N，当它满足以下条件的时候返回 true： 原数字旋转 180° 以后可以得到新的数字。 如 0, 1, 6, 8, 9 旋转 180° 以后，得到了新的数字 0, 1, 9, 8, 6 。 2, 3, 4, 5, 7 旋转 180° 后，得到的不是数字。 易混淆数 (confusing number) 在旋转180°以后，可以得到和原来不同的数，且新数字的每一位都是有效的。

解题方法

字典

使用字典保存每个可以翻转的字符翻转后会变成谁，然后对每一位数字进行翻转，看翻转后的数字和原来的数字是否相等。

C++代码如下：

class Solution {
public:
    bool confusingNumber(int N) {
        unordered_map<int, int> m{{0, 0}, {1, 1}, {6, 9}, {8, 8}, {9, 6}};
        int rotate = 0;
        int temp = N;
        while (temp != 0) {
            int mod = temp % 10;
            if (!m.count(mod))
                return false;
            rotate = 10 * rotate + m[mod];
            temp /= 10;
        }
        return rotate != N;
    }
};

日期

2019 年 9 月 18 日 —— 今日又是九一八