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…

258. 各位相加 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. 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? 找规律.假设 num = 384 = 3 * 100 + 8 * 10 + 4 第一轮计算 sum = 15 = 3 + 8 + 4 差值 = 3 * 99 + 8 * 9 = (3 *…

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…

258. 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…

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. 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. AC代码: class Solution(object): def addDigits(self, num): """ :type num: int :rtype: int """ num = str(num) num_len = le…

问题如下: 给一个非负整数 num,反复添加所有的数字,直到结果只有一个数字. 例如: 设定 num = 38,过程就像: 3 + 8 = 11, 1 + 1 = 2. 由于 2 只有1个数字,所以返回它. 进阶: 你可以不用任何的循环或者递归算法,在 O(1) 的时间内解决这个问题么? 初始的想法: 开始只看到了进阶,要求使用O(1)的时间复杂度,因此我想了一下,既然是int型变量,那么它的范围是-32768~32767,因此最高一共有5位数,所以O(1)算法可以直接使用五个int型变量存储起…

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. 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. class Solution: def addDigits(self, num: int) -> int: def hanshu(nums): sum = ): ge = nums % sum += ge nums = ) return sum sum = hanshu(num) ):…

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. 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…

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 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 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…

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…

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…

Leetcode 第 2 题(Add Two Numbers) 题目例如以下: Question You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linke…

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…

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 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.…

LeetCode:字符串相加[415] 题目描述 给定两个字符串形式的非负整数 num1 和num2 ,计算它们的和. 注意: num1 和num2 的长度都小于 5100.num1 和num2 都只包含数字 0-9.num1 和num2 都不包含任何前导零.你不能使用任何內建 BigInteger 库, 也不能直接将输入的字符串转换为整数形式. 题目分析 这道题其实很简单,我们要搞清楚手工计算两数之和的流程.两数相加,和如果大于10的话就有进位,进位最高为1,默认为0,该位相加的和应为sum%…

[LeetCode]738. Monotone Increasing Digits 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/monotone-increasing-digits/description/ 题目描述: Given a non-negative integer N, find th…

[LeetCode]402. Remove K Digits 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/remove-k-digits/description/ 题目描述: Given a non-negative integer num represented as a string, rem…

[LeetCode]423. Reconstruct Original Digits from English 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/reconstruct-original-digits-from-english/description/ 题目描述: Given a non-empty string containing an out-of-order English representatio…

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…

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…

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…

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…