Palindrome-number

最新推荐文章于 2024-01-19 11:14:01 发布

转载最新推荐文章于 2024-01-19 11:14:01 发布 · 234 阅读

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。通过反转整数的一部分并与原始数字对比来实现，有效避免了溢出问题。

题目描述

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 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.

code:
public class Solution {
    public boolean isPalindrome(int x) {
        if(x < 0 || x != 0 && x % 10 == 0)
            return false;
        int s = 0;
        while(s < x)
        {
            s = s * 10 + x % 10;
            x = x / 10;
        }
        return (s == x || s / 10 == x);
    }
}

当数字个数是奇数时，最中间的数会在最后一次循环中给了reverse，所以reverse/10去掉最后一个数，另外x的值要的就是其中前一半的数字，不需要保存,

不需要考虑溢出问题，因为原来的数位5位，用到的辅助数只有原数的一半长度2或3位，原数没有溢出，那么用到的数就不会溢出。还有就是本函数只是判断是否是回文，而没有输出回文