LeetCode - Palindrome Number

最新推荐文章于 2025-09-07 20:32:33 发布

weixin_33946020

最新推荐文章于 2025-09-07 20:32:33 发布

阅读量57

点赞数

文章标签：数据结构与算法

本文介绍了一种不使用额外空间判断整数是否为回文数的方法，提供了两种解决方案：一种是直接反转整数进行比较；另一种是通过计算10的幂次来逐位比较。

题目：

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.

思路：

两种解法，第一种就是reverse它，有溢出也不怕；第二种就是把跟它位数相同的10的最大次方求出来，然后同时除同时余，对比两边。

package manipulation;

public class PalindromeNumber {
    
    public boolean isPalindrome(int x) {
        if (x < 0) return false;
        int y = x;
        int result = 0;
        while (y > 0) {            
            result = result * 10 + y % 10;            
            y = y / 10;
        }
        
        return x == result;
    }
    
    public boolean isPalindrome2(int x) {
        if (x < 0) return false;
        int a = 1;
        while (x / a >= 10) {
            a = a * 10;
        }
        
        int left = 0;
        int right = 0;
        while (x > 0 && a > 1) {
            left = x / a;
            right = x % 10;
            if (left != right) return false;
            x = (x % a) / 10;
            a = a / 100;
        }
        
        return true;
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        PalindromeNumber p = new PalindromeNumber();
        System.out.println(p.isPalindrome(1234567890));
    }

}