# 206-reverse-linked-list


Try it on <a href='https://leetcode.com/problems/206-reverse-linked-list'>leetcode</a>

## Description
<div class="description">
<div><p>Given the <code>head</code> of a singly linked list, reverse the list, and return <em>the reversed list</em>.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/rev1ex1.jpg" style="width: 542px; height: 222px;">
<pre><strong>Input:</strong> head = [1,2,3,4,5]
<strong>Output:</strong> [5,4,3,2,1]
</pre>

<p><strong>Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/rev1ex2.jpg" style="width: 182px; height: 222px;">
<pre><strong>Input:</strong> head = [1,2]
<strong>Output:</strong> [2,1]
</pre>

<p><strong>Example 3:</strong></p>

<pre><strong>Input:</strong> head = []
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is the range <code>[0, 5000]</code>.</li>
	<li><code>-5000 &lt;= Node.val &lt;= 5000</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> A linked list can be reversed either iteratively or recursively. Could you implement both?</p>
</div>
</div>

## Solution(Python)
```Python
# 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
        cur = head

        while cur:
            _next = cur.next
            cur.next = prev
            prev = cur
            cur = _next

        return prev

```