Let Code Day 30 starting from scratch "234. Palindrome Linked List"

Overview

It seems that coding tests are conducted overseas in interviews with engineers, and in many cases, the main thing is to implement specific functions and classes according to the theme.

As a countermeasure, it seems that a site called Let Code will take measures.

A site that trains algorithmic power that can withstand coding tests that are often done in the home.

I think it's better to have the algorithm power of a human being, so I'll solve the problem irregularly and write down the method I thought at that time as a memo.

Leetcode

Leet Code Table of Contents Starting from Zero

Last time Leet Code Day 29 "46. Permutations" starting from zero

Basically, I would like to solve the easy acceptance in descending order.

Twitter I'm doing it.

problem

234. Palindrome Linked List

You will be given a unidirectional linked list, so check if it is a palindrome.

Example 1:

Input: 1->2 Output: false

Example 2:

Input: 1->2->2->1 Output: true

solution

The easiest way is to return True if it matches the flipped list of the given linked list, or False if it doesn't.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def isPalindrome(self, head: ListNode) -> bool:
        value = []
        while head:
            value.append(head.val)
            head = head.next
        return value == value[::-1]
# Runtime: 92 ms, faster than 15.28% of Python3 online submissions for Palindrome Linked List.
# Memory Usage: 24.1 MB, less than 11.54% of Python3 online submissions for Palindrome Linked List.

I didn't have a little time this time, so I'll bring someone else's answer from discuss.

For example, the answer to O (1).

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def isPalindrome(self, head):
        rev = None
        fast = head
        while fast and fast.next:
            fast = fast.next.next
            rev, rev.next, head = head, rev, head.next
        tail = head.next if fast else head
        isPali = True
        while rev:
            isPali = isPali and rev.val == tail.val
            head, head.next, rev = rev, head, rev.next
            tail = tail.next
        return isPali
# Runtime: 92 ms, faster than 15.28% of Python3 online submissions for Palindrome Linked List.
# Memory Usage: 24.1 MB, less than 11.54% of Python3 online submissions for Palindrome Linked List.

https://leetcode.com/problems/palindrome-linked-list/discuss/64500/11-lines-12-with-restore-O(n)-time-O(1)-space

The contents of this discussion are recommended as an easy-to-understand explanation. It's amazing that it's easy to understand visually ...

https://leetcode.com/problems/palindrome-linked-list/discuss/324358/O(n)-time-and-O(1)-space-with-explanation-(Python-and-C)

Later, there were multiple solutions in the linked list chapter of this book.

150 questions to train your programming skills to fight in the world

If you have time, you may want to check it out.