小白笔记----------------------leetcode(9. Palindrome Number)

最新推荐文章于 2024-07-11 15:45:00 发布

Aaron92

最新推荐文章于 2024-07-11 15:45:00 发布

阅读量360

点赞数

CC 4.0 BY-SA版权

分类专栏：算法设计

本文链接：https://blog.youkuaiyun.com/Aaron92/article/details/53884428

算法设计专栏收录该内容

56 篇文章

订阅专栏

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。通过数学运算实现整数反转并比较，解决了负数及整数溢出的问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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?

这里要求不能用多余空间，那就仅仅依靠/ 和%这两个运算就能够成功地将这个题解出来了

代码：

public class Solution {
    public boolean isPalindrome(int x) {
        if(x < 0){
            return false;
        }else if(x < 10){
            return true;
        }
        int i = 1;
        int s = x;
        int r = x;
        int reverse = 0,t;
        boolean flag;
        while(x >= 10){
            x = x/10;
            i++;
        }
        int j = 0;
        while(i > 1 && j < i){
            reverse = reverse * 10 + s%10;
            s = s/10;
            j++;
        }
        if(reverse == r){
            flag = true;
        }else{
            flag = false;
        }
        return flag;
        
    }
}