palindrome-number

/**
* @author gentleKay
* 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.
*
* 确定整数是否为回文。这样做没有额外的空间。
* 单击以显示扰流器。
* 一些提示：
* 负整数可以是回文吗？（即-1）
* 如果要将整数转换为字符串，请注意使用额外空间的限制。
* 您还可以尝试反转一个整数。但是，如果您解决了“反向整数”问题，
* 您就会知道反向整数可能溢出。你将如何处理这样的案件？
* 有一种更通用的方法来解决这个问题。
*/

/**
 * @author gentleKay
 * 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.
 * 
 * 确定整数是否为回文。这样做没有额外的空间。
 * 单击以显示扰流器。
 * 一些提示：
 * 负整数可以是回文吗？（即-1）
 * 如果要将整数转换为字符串，请注意使用额外空间的限制。
 * 您还可以尝试反转一个整数。但是，如果您解决了“反向整数”问题，
 * 您就会知道反向整数可能溢出。你将如何处理这样的案件？
 * 有一种更通用的方法来解决这个问题。
 */

public class Main12 {
	public static void main(String[] args) {
		int x = 123421;
		System.out.println(Main12.isPalindrome(x));
	}
	
	public static boolean isPalindrome(int x) {
		if (x < 0) {
			return false;
		}
		int last = 0;
		int res = x;
		while (res != 0) {
			last = last*10 + res%10;
			res = res / 10;
		}
		return last == x;
    }
}