## Description

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

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 the range
`[0, 5000]`

. `-5000 <= Node.val <= 5000`

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

## Explanation

Keep track of the previous nodes. Attach each node’s next to the previous node.

## Python Solution

```
# 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]:
prev = None
while head != None:
temp = head.next
head.next = prev
prev = head
head = temp
return prev
```

- Time complexity: O(N).
- Space complexity: O(1).

## One Thought to “LeetCode 206. Reverse Linked List”