9. 回文数 @Python @Go

原题

https://leetcode.cn/problems/palindrome-number/description/

思路

循环

复杂度

时间：O(len(n))
空间：O(len(n))

Python代码

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if x < 0:
            return False
        p = 0 
        prev = x
        while x: 
            digit = x % 10
            p = 10 * p + digit
            x //= 10 
        return p == prev
        

Go代码

func isPalindrome(x int) bool {
    if x < 0 {
        return false
    }
    // 计算回文数
    var y int = 0
    var z int = x
    for z != 0 {
        y = 10 * y + z % 10
        z /= 10
    }
    return x == y
}