LeetCode_Reverse Integer

Posted on | Edited on | In | Views:

Reverse Integer

Given a 32-bit signed integer, reverse digits of an integer.
（翻转一个有符号整形数值字符串）

Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: \([−2^{31}, 2^{31} − 1]\). For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Example:



1. 字符串翻转

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
class Solution:
    def reverse(self, x):
        """
        :type x: int
        :rtype: int
        """
        string = str(x)
        n = len(string)
        
        if string[0] == '-':
            result = [string[n-i] for i in range(1, n)]
            result = ['-'] + result
        else:
            result = [string[n-i-1] for i in range(n)]
        
        re = int("".join(result))
        
        if re > pow(2, 31) - 1 or re < - pow(2, 31):
            re = 0
        return re

注：超过有符号整形的范围时需要返回0。另外string[::-1]也可以直接实现字符串的翻转。

2. 整形翻转

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
class Solution:
    def reverse(self, x):
        """
        :type x: int
        :rtype: int
        """
        
        abso_x = abs(x)
        result = 0
        
        while abso_x != 0 :
            a = abso_x % 10
            abso_x = abso_x // 10
            result = result * 10  + a
            
        if x < 0:
            result *= -1  
            
        if result > (pow(2, 31) - 1) or result < -pow(2, 31):
            return 0
        
        return result