palindrome-number

packagecom.ytx.array;

/** 题目:palindrome-number

 *

 *  描述：

 *     Determine whether an integer is apalindrome. Do this without extra space.

             click to show spoilers.

             Some hints:

             Could negative integers bepalindromes? (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.

 *

 *@authoryuantian xin

 *

 * 判断一个数字是不是回文数字，不允许申请额外的存储空间。

 *

 * 思路：回文数字就是从头读到尾和从尾读到头是一样，比如1112111,3663,233332。那么我把数x倒置后得到reverse，

 * 判断它们是否相等即可,但是以下程序类似10,100,1000等10的倍数，其实得到的倒置reverse是不正确的，但是不影响

 * 最终结果，因为它们肯定不是回文数字，还有负数也肯定不是。

 *

 */

publicclass Palindrome_number {

       

       publicbooleanisPalindrome(intx) {

             

             intreverse = 0;

             inttemp = x;

          while(temp > 0) {

              reverse= reverse * 10 + temp % 10 ;

              temp= temp / 10;

           }

       return (x==reverse);

    }

       publicstatic void main(String[]args) {

             

             System.out.println(newPalindrome_number().isPalindrome(1112111));

       }

}