此篇文章為我的解題紀錄，程式碼或許並不是很完善

Leetcode - 9. Palindrome Number

解題思路

將x轉為字串後再跑迴圈檢查是否為回文

我滴程式碼

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class Solution:
    def isPalindrome(self, x: int) -> bool:
        word = str(x)
        limit = int(len(word)/2)
        size = len(word)
        for i in range(limit):
            if (word[i] != word[size - i - 1]):
                return False
        return True

看過大神後的程式碼

將字串翻轉後再檢查兩個字串是否相等

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class Solution:
    def isPalindrome(self, x: int) -> bool:
        word = str(x)
        txt = word[::-1]
        if word == txt:
            return True 
        return False