9. Palindrome Number

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

 方法一:

思路: 判断一个整数是否为回文结构与Reverse Integer思路几乎相同,稍微修改即可。

 1 class Solution(object):
 2     def isPalindrome(self, x):
 3         """
 4         :type x: int
 5         :rtype: bool
 6         """
 7         list = []
 8         a = abs(x)
 9         y = 0
10         for i in range(len(str(a))):
11             i = a % 10
12             list.append(i)
13             a = a // 10
14 
15         list.reverse()
16        
17         for j in range(len(list)):
18             y = y + list[j]* 10 ** j
19         if x < 0:
20            y = 0 - y
21 
22         return 0<=x==y<2147483648

 方法二:

思路:将x转化为字符串再翻转,再将翻转后的字符串转化为整数,判断两个整数是否相等。

 1 class Solution(object):
 2     def isPalindrome(self, x):
 3         """
 4         :type x: int
 5         :rtype: bool
 6         """
 7         if x < 0:
 8             return False
 9         else:
10             r = int(str(x)[::-1])
11             return x == r
原文地址:https://www.cnblogs.com/caowenhao/p/8423174.html