258. Add Digits 入学考试:数位相加】的更多相关文章

[抄题]: 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. [暴力解法]: 时间分析: 空间…
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…
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…
[抄题]: 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. [暴力解法]: 时间分析: 空间分析: [优化后]: 时间分析…
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…
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? 详见:https://leetcode.com/problems/add-digits/description/ Java实现: class Solution { public int addDigits(int num) { w…
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…