[LEET CODE] 206. Reverse Linked List

# Problem

Given the head of a singly linked list, reverse the list, and return the reversed list. Constraints: The number of nodes in the list is the range [0, 5000]. -5000 <= Node.val <= 5000

 

# My Answer

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        node = head
        rev = None
        while node and node.next:
            rev, rev.next, node = node, rev, node.next
        if node:
            rev,rev.next = node,rev
        return rev

 

# Solution 1 - 재귀 구조로 뒤집기

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        def reverse(node: ListNode, prev: ListNode = None):
            if not node:
                return prev
            next, node.next = node.next, prev
            return reverse(next, node)

        return reverse(head)

 

# Solution 2 - 반복 구조로 뒤집기

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        node, prev = head, None

        while node:
            next, node.next = node.next, prev
            prev, node = node, next

        return prev