X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X.  Each digit must be rotated - we cannot choose to leave it alone. A number is valid if each digit remains a digit after rotat…

［抄题］: X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X.  Each digit must be rotated - we cannot choose to leave it alone. A number is valid if each digit remains a digit after…

X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X.  Each digit must be rotated - we cannot choose to leave it alone. A number is valid if each digit remains a digit after rotat…

题目要求 X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X.  Each digit must be rotated - we cannot choose to leave it alone. A number is valid if each digit remains a digit after…

https://leetcode.com/problems/rotated-digits/ X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X.  Each digit must be rotated - we cannot choose to leave it alone. A number is v…

我们称一个数 X 为好数, 如果它的每位数字逐个地被旋转 180 度后,我们仍可以得到一个有效的,且和 X 不同的数.要求每位数字都要被旋转. 如果一个数的每位数字被旋转以后仍然还是一个数字, 则这个数是有效的.0, 1, 和 8 被旋转后仍然是它们自己:2 和 5 可以互相旋转成对方:6 和 9 同理,除了这些以外其他的数字旋转以后都不再是有效的数字. 现在我们有一个正整数 N, 计算从 1 到 N 中有多少个数 X 是好数? 示例: 输入: 10 输出: 4 解释: 在[1, 10]中有四个…

problem 788. Rotated Digits solution1: class Solution { public: int rotatedDigits(int N) { ; ; i<=N; ++i) { if(check(i)) res++; } return res; } bool check(int num) { string str = to_string(num); bool flag = false; for(auto ch:str) { ') return false;…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/rotated-digits/description/ 题目描述 X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that i…

X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X.  Each digit must be rotated - we cannot choose to leave it alone. A number is valid if each digit remains a digit after rotat…

翻译 给定一个非负整型数字,反复相加其全部的数字直到最后的结果仅仅有一位数. 比如: 给定sum = 38,这个过程就像是:3 + 8 = 11.1 + 1 = 2.由于2仅仅有一位数.所以返回它. 紧接着: 你能够不用循环或递归在O(1)时间内完毕它吗? 原文 Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. For example: Give…