Leetcode算法——7、翻转整数

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题目

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

给定一个32位的有符号整数，将其所有数字反转。
备注：
假设只处理[−2^31, 2^31 − 1]范围内的整数。当反转后的结果溢出时，返回0。

示例：
Example 1:
Input: 123
Output: 321

Example 2:
Input: -123
Output: -321

Example 3:
Input: 120
Output: 21

思路

这道题比较简单，定义一个新的变量，每次将原数的最后一位提取出来，放入到新数的最后一位即可。这样，原数越来越短，新数越来越长，直至原数长度为0。

python代码

def reverse(x):
    """
    :type x: int
    :rtype: int
    """

    result = 0
    a = abs(x)
    while(a != 0):
        pop = a % 10
        a = (a - pop) / 10
        result = result * 10 + pop
    if x > 0 and result < 2**31:
        return int(result)
    if x < 0 and result <= 2**31:
        return -int(result)
    return 0

if '__main__' == __name__:
    x = -2147483648
    print(reverse(x))