question: 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. answer:(which don't think b…

Add Digits 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…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 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…

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 方法一:递归 方法二:减1模9 方法三:直接模9 日期 [LeetCode] 题目地址:https://leetcode.com/problems/add-digits/ Total Accepted: 33351 Total Submissions: 71187 Difficulty: Easy 题目描述 Given a non-negative…

题目: 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…

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…

这是悦乐书的第199次更新,第207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63题(顺位题号是258).给定非负整数num,重复添加其所有数字,直到结果只有一位数.例如: 输入:38 输出:2 说明:过程如下:3 + 8 = 11,1 + 1 = 2.由于2只有一位数,所以请将其返回. 跟进:你可以在O(1)运行时间内没有任何循环/递归的情况下执行此操作吗? 本次解题使用的开发工具是eclipse,jdk使用的版本是1.8,环境是win7 64位系统,使用J…

一.题目描写叙述 Given two binary strings, return their sum (also a binary string). For example, a = "11" b = "1" Return "100". 二.解题技巧 这道题考察两个二进制数相加,考虑到输入的是两组string,同一时候注意在运算时从左到右各自是从低位到高位,因此须要考虑对输入进行翻转处理,中间二进制树相加部分没有过多的设计障碍.主要是计算进位:…

1.问题描述 2.问题分析 循环拆分数字,然求和判断. 3.代码 int addDigits(int num) { ) return num; int result = num; do{ vector<int> r = splitnum( result ); result = ; for(auto & n : r ){ result += n; } } ); return result ; } vector<int> splitnum( int num ){ vector&…

1.题目名称 Add Digits (非负整数各位相加) 2.题目地址 https://leetcode.com/problems/add-digits/ 3.题目内容 英文:Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. 中文:有一个非负整数num,重复这样的操作:对该数字的各位数字求和,对这个和的各位数字再求和……直到最后得到一个仅1位的数…