文章目录

• 题目描述
• 自己的解法
• 参考答案
• • Early Stop
• 总结
• Reference

题目描述

Given an integer x, return true if x is palindrome integer.

An integer is a palindrome when it reads the same backward as forward.

For example, 121 is a palindrome while 123 is not.

Example 1:

Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.

Example 2:

Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Constraints:

-231 <= x <= 231 - 1

Follow up: Could you solve it without converting the integer to a string?

自己的解法

I solved this problem with str and numerical operation.

class Solution:
    def isPalindrome(self, x: int) -> bool:
        # 1. str解法
        x_str = str(x)
        return x_str[::-1] == x_str 

class Solution:
    def isPalindrome(self, x: int) -> bool:
        # 2. 数值解法
        running_x = x
        reverse_x = 0
        while running_x > 0:
            reverse_x = 10 * reverse_x + running_x % 10 
            running_x = running_x//10
        return reverse_x == x

The time complexity is $O(lo{g}_{10}(n))$ and the space complexity is $O(1)$.

参考答案

Early Stop

The answer updated my code by comparing only half of the numbers.
But still, they have the same complexity.

class Solution:
    def isPalindrome(self, x: int) -> bool:
        # 3.参考答案改进
        if x < 0 or (x % 10 ==0 and x != 0) :
            return False
        reverse_x = 0
        while  reverse_x < x:
            reverse_x = reverse_x * 10 + x % 10 
            x = x//10
        return reverse_x == x or reverse_x//10 == x

Time complexity: $O(lo{g}_{10}(n))$
Space complexity: $O(1)$

总结

1. 锻炼了数值运算的一些小trick
2. Early Stop优化代码，思路++

Reference

https://leetcode.com/problems/palindrome-number/