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…

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 rotati…

788. 旋转数字 788. Rotated Digits 题目描述 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…

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;…

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…

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…

这是悦乐书的第316次更新,第337篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第185题(顺位题号是788).如果一个数字经过180度旋转后,变成了一个与原数字不同的数,这样的数被称为好数字.数字中的每一位都必须经过旋转.旋转的规则是:0,1,8这三个数旋转后还是自身,2旋转后变为5,5旋转后变为2,6旋转后变为9,9旋转后变为6,剩下的3,4,7旋转后并不能转成其他数字,是无效的.现在给出正数N,从1到N的好数字一共有多少个?例如: 输入:10 输出:4 说明…

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…

问题 该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/3967 访问. 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…

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…

1.题目描述 2.代码 int rotatedDigits(int N) { ; ; i <= N; i++) { if (isGood(i)) { res++; } } return res; } bool isGood(int x) { int rx = x; vector<int> nums; while (x) { ; nums.push_back(r); x = x / ; } for (vector<int>::iterator it = nums.begin()…

X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X. A number is valid if each digit remains a digit after rotation. 0, 1, and 8 rotate to themselves; 2 and 5 rotate to each othe…

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. For example: Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it. Follow up: Could you do it without an…

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…

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…

LeetCode上牵扯到Rotated Sorted Array问题一共有四题,主要是求旋转数组的固定值或者最小值,都是考察二分查找的相关知识.在做二分查找有关的题目时,需要特别注重边界条件和跳出条件.在做题的过程中,不妨多设几个测试案例,自己判断一下. 下面是这四题的具体解答. 33.Search in Rotated Sorted Array 在求旋转数组中查找固定值,数组中每个数唯一出现.返回查找索引. class Solution: def search(self, nums: List…

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. For example: Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it. Follow up:Could you do it without any…

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. For example: Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it. Follow up:Could you do it without any…

Description: Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. For example: Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it. Follow up:Could you do i…

［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…

题意: 将一个整数num变成它的所有十进制位的和,重复操作,直到num的位数为1,返回num. 思路: 注意到答案的范围是在区间[0,9]的自然数,而仅当num=0才可能答案为0. 规律在于随着所给自然数num的递增,结果也是在1~9内循环递增的,那么结果为(num-1)%9+1. C++: class Solution { public: int addDigits(int num) { ; )%+; } }; AC代码 python: class Solution(object): def…

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…

请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift    说明:题目中含有$符号则为付费题目. 如:[Swift]LeetCode156.二叉树的上下颠倒 $ Binary Tree Upside Down 请下拉滚动条查看最新 Weekly Contest!!! Swift LeetCode 目录 | Catalog 序        号 题名Title 难度     Difficulty  两数之…

突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对应的随笔下面评论区留言,我会及时处理,在此谢过了. 过程或许会很漫长,也很痛苦,慢慢来吧. 编号 题名 过题率 难度 1 Two Sum 0.376 Easy 2 Add Two Numbers 0.285 Medium 3 Longest Substring Without Repeating C…

645. Set Mismatch The set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another nu…

All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017) .   Top Interview Questions # Title Difficulty Acceptance 1 Two Sum Medium 17.70% 2 Add Two N…

终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 如果各位看官们,大神们发现了任何错误,或是代码无法通过OJ,或是有更好的解法,或是有任何疑问,意见和建议的话,请一定要在对应的帖子下面评论区留言告知博主啊(如果不方便注册博客园的话,可以下载下文提到的APP,在Feedback中给博主发邮件交流哈),同时也请大家踊跃地,大量地,盲目地提供各个题目的follow up一起讨论哈,多谢多谢,祝大家刷得愉快…

Contest 71 () Contest 72 () Contest 73 (2019年1月30日模拟) 链接:https://leetcode.com/contest/weekly-contest-73 结果:2/4,会做第一题和第三题,第二题和第四题不会做. [788]Rotated Digits(第一题 4分)(谷歌tag) 给了一个 good number 的定义,X is a good number if after rotating each digit individually…

字符串篇 # 题名 刷题 通过率 难度 3 无重复字符的最长子串   24.6% 中等 5 最长回文子串   22.4% 中等 6 Z字形变换   35.8% 中等 8 字符串转整数 (atoi)   15.3% 中等 10 正则表达式匹配   18.4% 困难 12 整数转罗马数字   53.8% 中等 13 罗马数字转整数 C#LeetCode刷题之#13-罗马数字转整数(Roman to Integer) 53.7% 简单 14 最长公共前缀 C#LeetCode刷题之#14-最长公共前缀…