9. Palindrome Number

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本文介绍了一种不使用额外空间判断整数是否为回文的方法。通过特定逻辑判断特殊情况，并利用循环逐步比较数字前后部分来实现。文章还讨论了负数、整数溢出等问题。

Determine whether an integer is a palindrome. Do this without extra space.

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.

public class Solution {
    public boolean isPalindrome(int x) {
        if(x<0)
			return false;
		if(x==0)
			return true;
		if(x%10==0)
			return false;
		if(x<10)
			return true;
		if(x<100 && x%11==0)
			return true;
		if(x<1000 && (x/100==x%10))
			return true;
                //对一些简单情况进行剪枝
		int num=0;
		while(x > num){
			num = num*10 + x%10;
			x = x/10;
		}
		return (x==num || x==num/10);
               //把数字后半部分一位一位的移到另一个数，然后比较。偶数位则直接相比，奇数位则要忽略一位。
 }
}