【LeetCode】1056. Confusing Number 解题报告(C++)
作者:互联网
- 作者: 负雪明烛
- id: fuxuemingzhu
- 个人博客:http://fuxuemingzhu.cn/
目录
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:
0 <= N <= 10^9
- 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
标签:11,rotate,1056,after,Number,C++,number,180,confusing 来源: https://blog.csdn.net/fuxuemingzhu/article/details/100992424