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…

题目链接:https://leetcode-cn.com/problems/add-digits/ 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? 常规思路: int addDigits(int x) { ) return x; ; while(x){ s…

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…

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…

本文出处:http://www.cnblogs.com/wy123/p/6217772.html 字符串自身相加, 虽然赋值给了varchar(max)类型的变量,在某些特殊情况下仍然会被“截断”,这到底是varchar(max)长度的问题还是操作的问题? 1,两个不超过8000长度的字符串自身相加,其结果长度超过8000之后会被截断: 不多说,直接上例子:定义一个字符串,赋值给 varchar(max)类型的变了,字符创长度为4040没有,任何问题. 把4040长度的字符串复制一份出来,也就是…

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.…

Given two binary strings, return their sum (also a binary string). For example,a = "11"b = "1"Return "100". 二进制数想加,并且保存在string中,要注意的是如何将string和int之间互相转换,并且每位相加时,会有进位的可能,会影响之后相加的结果.而且两个输入string的长度也可能会不同.这时我们需要新建一个string,它的长度是两…

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 linked list. Input: (2 -> 4 -> 3) + (5 -> 6 ->…

#include<stdio.h>#include<string.h>void main(){ int a; int b; char str1[10] = "99999"; char str2[10] = "1111111"; char str[30]; int k = 0, i = 0, j = 0; for (k = 0; k < 30&&i<strlen(str1);){ str[k++] = str1[i+…

18.1 Write a function that adds two numbers. You should not use + or any arithmetic operators. 这道题让我们实现两数相加,但是不能用加号或者其他什么数学运算符号,那么我们只能回归计算机运算的本质,位操作Bit Manipulation,我们在做加法运算的时候,每位相加之后可能会有进位Carry产生,然后在下一位计算时需要加上进位一起运算,那么我们能不能将两部分拆开呢,我们来看一个例子759+674 1.…