206. Reverse Linked List

https://leetcode.com/problems/reverse-linked-list/

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function(head) {
  if (!head) {
    return head
  }

  var current = head.next
  head.next = null

  while (current) {
    let next = current.next
    current.next = head
    head = current
    current = next
  }

  return head
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function(head) {
  if (!head || !head.next) {
    return head
  }

  var next = head.next
  var reverse = reverseList(next)
  next.next = head
  head.next = null

  return reverse
}

Python

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


class Solution(object):

    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head:
            return head
        current = head.next
        head.next = None
        while current:
            nxt = current.next
            current.next = head
            head, current = current, nxt
        return head

Python

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


class Solution(object):

    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
            return head
        nxt, reverse = head.next, self.reverseList(head.next)
        nxt.next, head.next = head, None
        return reverse