[Leetcode]Reverse Integer@python

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output:  321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:

Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

整数反转

解法一：将int整数变成字符串，按list进行翻转，然后再转成int

class Solution(object):
def reverse(self, x):
"""
:type x: int
:rtype: int
"""
flag = 0
if x<0:
flag = 1
res = ""
x = str(abs(x))
for i in range(1,len(x)+1):
res += x[-i]
if int(res) > 2147483648:
return 0
if flag == 0:
return int(res)
else:
return -int(res)

解法二：先把数字求绝对值x，然后x%10取最后一位，然后ans = ans*10 + x%10，加上最后一位。然后x去掉最后一位。知道x = 0.要注意的时候，超出32位int类型的时候设置等于0就行了。

class Solution:
# @return an integer
def reverse(self, x):
answer = 0
sign = 1 if x > 0 else -1
x = abs(x)
while x > 0:
answer = answer * 10 + x % 10
x /= 10
return sign*answer