翻转数字

Question: 翻转数字

👉LeetCode链接👈

Reverse digits of an integer.

Have you thought about this?

Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer’s last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.

这个问题是将一个 int 类型的数字翻转，但是翻转时要考虑几个问题，上面提到了：溢出，或者最后一位是 0 怎么办？对于溢出，给定了规则，溢出的话返回 0 。

Example

1
2
3
Example1: x = 123, return 321

Example2: x = -123, return -321

Answer

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
def reverse(self, x):
    """
    :type x: int
    :rtype: int
    """
    max = 2**31-1
    if x > max > 0 or x < -max < 0:
        return 0
    x = x.__str__()[::-1]
    if '-' in x:
        x = -int(x[0:-1])
        if x < -max:
            return 0
    elif int(x) > max > 0:
        return 0
    return int(x)

Finally

感觉自己的解法很蠢，可是Leetcode上还没有 Solution 如果有人知道好一点的解法的话希望能告诉我，谢谢！

Mail To：rayson951005@gmail.com