<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 class="example">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 class="example">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 class="example">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>

# 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]:
        # Initialize pointers
        prev = None  # Previous node starts as None
        curr = head  # Current node starts at the head

        # Traverse the list
        while curr is not None:
            next_node = curr.next  # Save the next node
            
            curr.next = prev  # Reverse the link
            
            # Move pointers forward
            prev = curr  # Move prev to the current node
            curr = next_node  # Move curr to the next node

        # prev is now the new head of the reversed list
        return prev
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