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 an…

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…

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…

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…

258. 各位相加 258. Add Digits 题目描述 给定一个非负整数 num,反复将各个位上的数字相加,直到结果为一位数. LeetCode258. Add Digits 示例: 输入: 38 输出: 2 解释: 各位相加的过程为: 3 + 8 = 11, 1 + 1 = 2. 由于 2 是一位数,所以返回 2. 进阶: 你可以不使用循环或者递归,且在 O(1) 时间复杂度内解决这个问题吗? Java 实现 class Solution { public int addDigits(i…

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…

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…

258. Add Digits Digit root 数根问题 /** * @param {number} num * @return {number} */ var addDigits = function(num) { var b = (num-1) % 9 + 1 ; return b; }; //之所以num要-1再+1;是因为特殊情况下:当num是9的倍数时,0+9的数字根和0的数字根不同. 性质说明 1.任何数加9的数字根还是它本身.(特殊情况num=0)        小学学加法的…

lc 258 Add Digits lc 258 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. F…

Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2. Note: The length of both num1 and num2 is < 5100. Both num1 and num2 contains only digits 0-9. Both num1 and num2 does not contain any leading zero.…

LeetCode--Add Strings Question Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2. Note: The length of both num1 and num2 is < 5100. Both num1 and num2 contains only digits 0-9. Both num1 and num2 doe…

258. Add Digits Easy Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. Example: Input: 38 Output: 2 Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2.   Since 2 has only one digit, return it. F…

最近做的题记录下. 258. 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. int addDigi…

258. 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…

Add Digits Total Accepted: 49702 Total Submissions: 104483 Difficulty: Easy 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.…

数学题 172. Factorial Trailing Zeroes Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. (Easy) 分析:求n的阶乘中末位0的个数,也就是求n!中因数5的个数(2比5多),简单思路是遍历一遍,对于每个数,以此除以5求其因数5的个数,但会超时. 考虑到一个数n比他小…

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…

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…

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…

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…

Problem: 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 w…

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…

题目描述: 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. 解题思路: 假设输入的数字是一个5位数字num,则num的各位…

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…

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 withou…

题目要求 Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. 题目分析及思路 给定一个非负整数,要求将它的各位数相加得到一个新数并对该新数重复进行这样的过程,直到最后的数只有一位.可以使用递归的方法,不断地将给定数的各位数相加,跳出递归的条件是该数只有一位. python代码 class Solution: def addDigits(self,…

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

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…