LeetCode解题报告--Palindrome Number

最新推荐文章于 2020-05-30 22:34:34 发布

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最新推荐文章于 2020-05-30 22:34:34 发布

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分类专栏： java 算法 LeetCode解题报告 LeetCode 解题报告文章标签： leetcode java 算法 code

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本文介绍如何使用Java代码来判断整数是否为回文数，包括处理负数和包含多个零的情况。

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题目：回文数字的判断
Determine whether an integer is a palindrome. Do this without extra space.

Some hints:
Could negative integers be palindromes? (ie, -1)

If you are thinking of converting the integer to string, note the restriction of using extra space.

You could also try reversing an integer. However, if you have solved the problem “Reverse Integer”, you know that the reversed integer might overflow. How would you handle such case?

There is a more generic way of solving this problem.

原题链接地址：https://leetcode.com/problems/palindrome-number/

分析：该题难度不大，题意是给定一个int型数字num，判断其是否为回文数字。主要有两种情况：
1. num >= 0, 当num = 00000, 含多个零时，要单独讨论；
2. num < 0, 先转为正数，再进行判断。

Java 代码：（accepted）


public class PalindromeNuber {
    public static boolean isPlindrome(int x){
        if(x == 0) {
            x = 0;
        }
        String x1 = "" + x;
        StringBuffer x2 = new StringBuffer(x1); //Save reversed x1;

        if(x1.equals(x2.reverse().toString()))
            return true;
        else {
            return false;
        }
    }

    public static void main(String[] args) {
        int x = -2147447412;
        System.out.println(isPlindrome(x));
        int x1 = -0000000;
        System.out.println(isPlindrome(x1));
        int x2 = 103423;
        System.out.println(isPlindrome(x2));
        int x3 = -2192;
        System.out.println(isPlindrome(x3));
        int x4 = 00000;
        System.out.println(isPlindrome(x4));
    }
}