笔者最近在一边看<JS高级程序设计3>一边在FCC上找题目练习啊.那叫一个爽.这不,刚刚用生命在课堂,寝室,实验室,图书馆等各种场所将第五章"引用类型"搞定,FCC便知趣的给笔者来了个"回文数",笔者咬牙切齿,花了两天时间,又是研究数组,又是研究字符串,又是研究作用域,还看了很长时间的正则表达式.还好,不负有心人,嘿嘿嘿,现在为大家详细分享用JS实现精准回文数的辨别!!! 先给大家看几个类型的字符串: race car not a palindrome…
Determine whether an integer is a palindrome. Do this without extra space. click to show spoilers. Some hints: Could negative integers be palindromes? (ie, -1) If you are thinking of converting the integer to string, note the restriction of using ext…
题目描述: Find the largest palindrome made from the product of two n-digit numbers. Since the result could be very large, you should return the largest palindrome mod 1337. Example: Input: 2 Output: 987 Explanation: 99 x 91 = 9009, 9009 % 1337 = 987 Note…
题目等级:Easy 题目描述: Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward. Example 1: Input: 121 Output: true Example 2: Input: -121 Output: false Explanation: From left to right, it reads -12…
Problem Description 一个正整数,如果从左向右读(称之为正序数)和从右向左读(称之为倒序数)是一样的,这样的数就叫回文数.任取一个正整数,如果不是回文数,将该数与他的倒序数相加,若其和不是回文数,则重复上述步骤,一直到获得回文数为止.例如:68变成154(68+86),再变成605(154+451),最后变成1111(605+506),而1111是回文数.于是有数学家提出一个猜想:不论开始是什么正整数,在经过有限次正序数和倒序数相加的步骤后,都会得到一个回文数.至今为止还不知道…
好久没写java的代码了, 今天闲来无事写段java的代码,算是为新的一年磨磨刀,开个头,算法是Java判断回文数算法简单实现,基本思想是利用字符串对应位置比较,如果所有可能位置都满足要求,则输入的是回文数,否则不是,不多说,上代码: import java.util.*; public class HiJava { public static void main(String[] args) { Scanner sc = new Scanner(System.in); System.out.p…
Ugly Problem Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0Special Judge Problem Description Everyone hates ugly problems. You are given a positive integer. You m…