[leetcode][math] Add Digits
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 = 111 + 1 = 2. Since 2 has
only one digit, return it.
Follow up:
Could you do it without any loop/recursion in O(1) runtime?
Hint:
- A naive implementation of the above process is trivial. Could you come up with other methods?
- What are all the possible results?
- How do they occur, periodically or randomly?
class Solution {
public:
int addDigits(int num) {
int res = 0;
bool doneFlag = false;
while(1){
res = 0;
while(num > 0){
res += num%10;
num /= 10;
}
if(res/10 == 0) break;
num = res;
}
return res;
}
};
class Solution {
public:
int addDigits(int num) {
if(num == 0) return 0;
if(num % 9 == 0) return 9;
return num%9;
}
};
[leetcode][math] Add Digits的更多相关文章
- LeetCode:Add Digits - 非负整数各位相加
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- ...
- 【LeetCode】Add Digits
Add Digits Given a non-negative integer num, repeatedly add all its digits until the result has only ...
- LN : leetcode 258 Add Digits
lc 258 Add Digits lc 258 Add Digits Given a non-negative integer num, repeatedly add all its digits ...
- LeetCode 258 Add Digits(数字相加,数字根)
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. ...
- [LeetCode] 258. Add Digits 加数字
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. ...
- LeetCode 258. Add Digits
Problem: Given a non-negative integer num, repeatedly add all its digits until the result has only o ...
- (easy)LeetCode 258.Add Digits
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. ...
- Java [Leetcode 258]Add Digits
题目描述: Given a non-negative integer num, repeatedly add all its digits until the result has only one ...
- LeetCode 258 Add Digits 解题报告
题目要求 Given a non-negative integer num, repeatedly add all its digits until the result has only one d ...
随机推荐
- ORA-01795: 列表中的最大表达式数为 1000
系统报出一SQL异常,内容如下: java.sql.SQLException: ORA-01795: maximum number of expressions in a list is 1000 找 ...
- DQL命令(查询)
select *或字段1,字段2... from 表名 [where 条件] 提示:*符号表示取表中所有列:没有where语句表示 查询表中所有记录:有wh ...
- open函数详解
转载:https://www.cnblogs.com/frank-yxs/p/5925574.html open函数用来在进程中打开文件,如果成功则返回一个文件描述符fd. ============= ...
- HTML5与后台服务器的数据流动问题
编辑中,尚未完稿...2017.7.14 1345 很多前端开发出来的HTML5可能对于后台开发者来说,并不是很清楚,也许像我一样一知半解.而且真的让人很糊涂的地方就是前端的JS如何与后端的数据库进行 ...
- spring boot约定优于配置的这种做法在如今越来越流行了
约定优于配置的这种做法在如今越来越流行了,它的特点是简单.快速.便捷.但是这是建立在程序员熟悉这些约定的前提上.而 Spring 拥有一个庞大的生态体系,刚开始转到 Spring Boot 完全舍弃 ...
- Android动态加载字节码
概述 面对App业务逻辑的频繁变更,如果每一次改变都对App进行一次升级,会降低App的用户体验,那么App进行模块化升级(这里与增量升级是不同的)是很好的解决方案,让用户在完全无感觉的情况下改变Ap ...
- Qt程序打包成exe可执行文件
很多Qt爱好者想发布自己的Qt软件,但却发现在其他没有安装Qt SDK的机器上无法运行,这就是本文想要说明的问题.现在网上大部分软件都要发布自己开发的应用程序,都会打包到exe文件中,待安装完exe文 ...
- OSGI项目中获取文件路径
假设想依据给定的文件名创建一个File实例,你可能会这么写: File file = new File(当前类.class.getResource("config").toURI( ...
- 解题报告 之 HDU5317 RGCDQ
解题报告 之 HDU5317 RGCDQ Description Mr. Hdu is interested in Greatest Common Divisor (GCD). He wants to ...
- Android之使用weight属性实现控件的按比例分配空间
从今天開始,把看书时候的知识点整理成博客, 这个比較简单,预计有经验的都用过,weight属性 在做Android布局的时候,常常遇到须要几个控件按比例分配空间的情况 比方下图效果 在底部设置两个bu ...