[leetcode]9. Palindrome Number

最新推荐文章于 2025-09-08 22:54:36 发布

转载最新推荐文章于 2025-09-08 22:54:36 发布 · 98 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/alfredsun/p/10809272.html

文章标签：

#runtime

博客提及一道简单题，若不将其转换为字符串处理，会牺牲一定的时间复杂度，还给出了转载链接。

简单题，如果不转string要牺牲一些时间复杂度：

Submission Detail

11509 / 11509 test cases passed.	Status: Accepted
Runtime: 128 ms Memory Usage: 13.2 MB	Submitted: 1 minute ago

class Solution:
    def isPalindrome(self, x: int) -> bool:
        # solve it without converting the integer to a string
        # deal 0
        if x==None:
            return False

        if x < 0:
            return False
        if x == 0:
            return True
        # noraml
        order = []
        while (True):
            remain = x % 10
            div = x // 10
            if div == 0 and remain ==0:
                break
            else:
                order.append(remain)
                x = x // 10

        # is Palindrome
        # notice lengthS >=
        lengthS = len(order)
        if lengthS % 2 == 1:
            # odd
            list1 = order[0:lengthS // 2 + 1]
            list2 = order[lengthS // 2:]
            list2.reverse()
            for index in range(len(list1)):
                if list1[index] == list2[index]:
                    if index == len(list1) -1:
                        return True
                else:
                    # no palindrome
                    return False
        else:
            # even
            list1 = order[0:lengthS // 2]
            list2 = order[lengthS // 2:]
            list2.reverse()
            for index in range(len(list1)):
                if list1[index] == list2[index]:
                    if index == len(list1)-1:
                        return True
                else:
                    # no palindrome
                    return False