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Given the head of a singly linked list, return true if it is a palindrome or false otherwise. Example 1:Input: head = [1,2,2,1]Output: trueExample 2:Input: head = [1,2]Output: false

Question

Given the head of a singly linked list, return true if it is a palindrome or false otherwise. Example 1:Input: head = [1,2,2,1]Output: trueExample 2:Input: head = [1,2]Output: false

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Solution

Para determinar si una lista enlazada es un palíndromo, podemos seguir estos pasos:

  1. Encontrar la mitad de la lista: Usaremos dos punteros, uno que avanza una posición a la vez (lento) y otro que avanza dos posiciones a la vez (rápido). Cuando el puntero rápido llegue al final, el puntero lento estará en la mitad de la lista.

  2. Invertir la segunda mitad de la lista: A partir del puntero lento, invertimos la segunda mitad de la lista.

  3. Comparar las dos mitades: Comparamos la primera mitad de la lista con la segunda mitad invertida. Si todas las posiciones coinciden, la lista es un palíndromo.

  4. Restaurar la lista (opcional): Si es necesario, podemos restaurar la lista a su estado original invirtiendo nuevamente la segunda mitad.

Aquí está el código que implementa estos pasos:

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def isPalindrome(head: ListNode) -> bool:
    if not head or not head.next:
        return True
    
    # Encontrar la mitad de la lista
    slow, fast = head, head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
    
    # Invertir la segunda mitad de la lista

This problem has been solved

Similar Questions

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