9. Palindrome Number(判断一个整数是否为回文数)_用palindrome1编写一个程序判断是否是回文数-优快云博客

本文介绍了一种不使用额外空间来判断整数是否为回文数的方法。文章提供了一个Java实现示例，通过反转整数并比较原数来确定其是否为回文，同时考虑了负数和溢出的问题。

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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?

题目大意

1.不能用额外的空间，不能转化为string在判断。
2.如果反转数字记得判断范围。

解题思路

1.反转数字在计算

AC代码（java）

package test;
class Solution {
    public boolean isPalindrome(int x) {
        if(x<0)return false;
        if(x<10)return true;
        if(x%10==0)return false;
        long   reverse = 0,tem = x;
        while(x!=0){
            reverse = reverse*10 + x%10;
            x /= 10;
        }
        if(reverse == tem){
            return true;
        }else{
            return false;
        }
    }
}
public class test {
public  static void main(String[] args){
    int a = 1221;
    Solution b = new Solution();
    if(b.isPalindrome(a)){
        System.out.println("yes!");
    }else{
        System.out.println("no!");
    }
}
}