0009. Easy|Palindrome Number.md

• 9 - Palindrome Number
• 难度：Easy｜ 简单
• 相关知识点： Math
• 题目链接：https://leetcode.com/problems/palindrome-number/description/

数学方法

Solution 1: 通过反转一半的数字来减少计算次数

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if x < 0 or (x % 10 == 0 and x != 0):
            return False

        reversed_half = 0
        while x > reversed_half:
            reversed_half = reversed_half * 10 + x % 10
            x //= 10

        # 处理数字位数为奇数的情况
        return x == reversed_half or x == reversed_half // 10

Solution 2：取整数反转后的值进行比较

def isPalindrome(x):
    if x < 0:
        return False
    original = x
    reversed_x = 0
    while x != 0:
        reversed_x = reversed_x * 10 + x % 10
        x //= 10
    return original == reversed_x

字符串

Solution 3: 转成字符串比较

def isPalindrome(x):     
  return str(x) == str(x)[::-1]

Solution 4：将整数转成数组进行比较

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if x < 0:
            return False

        str_x = str(x)
        left, right = 0, len(str_x) - 1

        while left < right:
            if str_x[left] != str_x[right]:
                return False
            left += 1
            right -= 1

        return True