Leetcode:(Python)Reverse Integer

題目連結：https://leetcode.com/problems/reverse-integer/description/

題意說明

• 給定一個數字，回傳反向數字
• 給定 456，回傳 654
• 給定-568，回傳 -865

初始想法

• 分離出正負
• 將數字轉為字串，反轉，再逐一結合
• 將正負和數字結合

實作程式碼

class Solution:
    def reverse(self, x: int) -> int:
        if x < 0:
            sign = '-'
        else:
            sign = '+'
        x = str(abs(x))

        number_string = ""
        for n in x[::-1]:
            number_string = number_string + n

        if int(sign+number_string) < 2**31-1 and int(sign+number_string) > -2**31:
            return int(sign+number_string)
        else:
            return 0

表現其差無比

反思一：不用使用迴圈反轉字串

其實不用使用迴圈，直接 x[::-1]就是反轉字串結果

class Solution:
    def reverse(self, x: int) -> int:
        if x < 0:
            sign = '-'
        else:
            sign = '+'
        x = str(abs(x))

        number_string = x[::-1]

        if int(sign+number_string) < 2**31-1 and int(sign+number_string) > -2**31:
            return int(sign+number_string)
        else:
            return 0

馬上快了5%

參考他人解法

比對上下界時，因為數字已經取絕對值，可以只比上界，然後再結合正負號

實作程式碼

class Solution:
    def reverse(self, x: int) -> int:
        if x < 0:
            sign = '-'
        else:
            sign = '+'
        x = str(abs(x))

        number_string = x[::-1]

        if int(number_string) < 2**31-1:
            return int(sign+number_string)
        else:
            return 0            

表現明顯上升

最後增加可讀性

class Solution:
    def reverse(self, x: int) -> int:
        if x < 0:
            sign = '-'
        else:
            sign = '+'
        number_string = str(abs(x))

        number_string_reverse = number_string[::-1]

        if int(number_string_reverse) < 2**31-1:
            return int(sign+number_string_reverse)
        else:
            return 0