---
title: 【LeetCode】0007. Reverse Integer
date: 2019-06-05
is_modified: false
disqus: cynthiahackmd
categories:
- "面試刷題"
tags:
- "LeetCode"
---

{%hackmd @CynthiaChuang/Github-Page-Theme %}
<br>

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

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**Example 1:**

Input: 123
Output: 321

**Example 2:**

Input: -123
Output: -321

**Example 3:**

Input: 120
Output: 21

<br><br>

> 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.

<br>

**Related Topics:** `Math`

## 解題邏輯與實作
這題是要翻轉整數，如果直接當字串用 slice 就可以輕鬆解決，在加上 python 幾乎沒有 overflow 的問題，所以過程中也不用特地處理，回傳做個判斷就好，就是要注意一下正負問題而已，喔...還有尾數為 0 的狀況...

class Solution:
   def reverse(self, x: int) -> int:
      if x == 0:
         return 0

      is_negative = x < 0
      x = x * -1 if is_negative else x 

      str_x = str(x).strip('0')
      str_x = str_x[::-1]
      x = int(str_x)
      
      x = x * -1 if is_negative else x            
      x = 0 if   x < -2147483648 or x > 2147483647 else x
      return x

<br>

不過這一題，應該不是希望這樣寫才對，所以又乖乖著墨了一版不轉成字串的版本。

class Solution:
   def reverse(self, x: int) -> int:
      if not x:
         return 0

      is_negative = x < 0
      x = x * -1 if is_negative else x 
      result = 0
      
      while x :
         result = result * 10 + x % 10
         x //= 10

      result = result * -1 if is_negative else result            
      result = 0 if result < -2147483648 or result > 2147483647 else result
      return result

## 其他連結
1. [【LeetCode】0000. 解題目錄](/x62skqpKStKMxRepJ6iqQQ)

<br><br>

> **本文作者**： 辛西亞．Cynthia
> **本文連結**： [辛西亞的技能樹](https://cynthiachuang.github.io/LeetCode-0007-Reverse-Integer) / [hackmd 版本](https://hackmd.io/@CynthiaChuang/LeetCode-0007-Reverse-Integer)
> **版權聲明**： 部落格中所有文章，均採用 [姓名標示-非商業性-相同方式分享 4.0 國際](https://creativecommons.org/licenses/by-nc-sa/4.0/deed.en) (CC BY-NC-SA 4.0) 許可協議。轉載請標明作者、連結與出處！