LeetCode : 7. Reverse Integerの解法

初めに

本日もLeetCodeやってまいります。
今回は以下の問題。

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-231, 231 - 1], then return 0.

Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

 

Example 1:

Input: x = 123
Output: 321
Example 2:

Input: x = -123
Output: -321
Example 3:

Input: x = 120
Output: 21
 

Constraints:

-231 <= x <= 231 - 1

回答

class Solution:
    def reverse(self, x: int) -> int:
        result_str = ""
        if x < 0:
            for digit in str(abs(x))[::-1]:
                if digit == 0:
                    continue
                result_str += digit
            if -2**31 > -int(result_str):
                return 0
            return -int(result_str)
        if x >= 0:
            for digit in str(x)[::-1]:
                if digit == 0:
                    continue
                result_str += digit
            
            if int(result_str) > 2**31 - 1:
                return 0
            return int(result_str)

これだとちょっと長いのでGPTにシンプルにしてもらいました笑

class Solution:
    def reverse(self, x: int) -> int:
        sign = -1 if x < 0 else 1
        s = str(abs(x))[::-1]
        result = sign * int(s)
        if result < -2**31 or result > 2**31 - 1:
            return 0
        return result

最後に

無駄なループ書きすぎてました涙
これからも精進します。