【LeetCode】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数的方法。通过将整数拆分成数字数组并比较首尾元素的方式，实现了高效判断。特别讨论了负数情况及整数反转可能带来的溢出问题。

9. Palindrome Number

Total Accepted: 117430 Total Submissions: 373389 Difficulty: Easy

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

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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.

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代码：

class Solution(object):
    def isPalindrome(self, x):
        """
        :type x: int
        :rtype: bool
        """
        if x<0:
            return 1==2
        array=[]
        while x!=0:
            num=x%10
            array.append(num)
            x=x/10
        length=len(array)
        i=0
        j=length-1
        while i<j:
            if array[i]==array[j]:
                i+=1
                j-=1
            else:
                break
        if i>=j:
            return 1==1
        else:
            return 1==2