206. Reverse Linked List

EasyLeetCode

Given the head of a singly linked list, reverse the list, and return the reversed list.

Example 1

Input: head = [1,2,3,4,5]

Output: [5,4,3,2,1]

Example 2

Input: head = [1,2]

Output: [2,1]

Example 3

Input: head = []

Output: []

Constraints:

• The number of nodes in the list is in the range [0, 5000]
• -5000 <= Node.val <= 5000

Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?

How to solve the problem

Code

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
# 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: Optional[ListNode]) -> Optional[ListNode]:
        cur = head
        pre = None
        while(cur!= None):
            temp = cur.next # Save the next element in previous linked list to temp
            cur.next = pre # let pre be the new next element in new linked list
            pre = cur # give cur to pre first
            cur = temp
        return pre # return head element

Complexity

• Time complexity: O(n)
• Space complexity: O(1)