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…

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…

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…

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…

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…

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…