Palindrome Number

问题来源：https://leetcode.com/problems/palindrome-number/

/** *  * <p> * ClassName PalindromeNumber * </p> * <p> * Description Determine whether an integer is a palindrome(n. 回文). 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. * </p> *  * @author TKPad [e-mail] *         <p> *         Date 2015年3月24日 上午10:48:32 *         </p> * @version V1.0.0 * */public class PalindromeNumber {    public boolean isPalindrome(int x) {        // 该判断在此题无作用，主要用来解决判断负数回文的情况        // 因为正数比负数少表示一位，所以当为Integer.MIN_VALUE的时候，是无法取绝对值转为正数的        if (x == Integer.MIN_VALUE) {            return false;        }        if (x < 0) {            // 这题说的不清晰，并没有明确说明负数是否可以是回文的，我认为如果不考虑符号的话，负数也可以是回文的，但是该题所提供的测试用例全部认为负数都是非回文的            return false;        }        // 我的解题思路是从低位开始，不断的逆向构造一个整数，那么该整数如果与要判断的数一样的话，那么肯定就是回文的        int temp = x;        int reverse_x = 0;        while (temp > 0) {            reverse_x = reverse_x * 10 + temp % 10;            temp /= 10;        }        return reverse_x == x;    }    public static void main(String[] args) {        // boolean palindrome = new PalindromeNumber().isPalindrome(2050880502);        boolean palindrome = new PalindromeNumber().isPalindrome(-205088050);        System.out.println(palindrome);    }}