LeetCode - Palindrome Number-优快云博客

本文讨论了如何不使用额外空间来判断一个整数是否为回文数，包括正负整数的情况，并提供了两种解决方法：反转整数和比较相同位数的整数部分。

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));
    }

}