<p>Given the <code>head</code> of a singly linked list, sort the list using <strong>insertion sort</strong>, and return <em>the sorted list&#39;s head</em>.</p>

<p>The steps of the <strong>insertion sort</strong> algorithm:</p>

<ol>
	<li>Insertion sort iterates, consuming one input element each repetition and growing a sorted output list.</li>
	<li>At each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list and inserts it there.</li>
	<li>It repeats until no input elements remain.</li>
</ol>

<p>The following is a graphical example of the insertion sort algorithm. The partially sorted list (black) initially contains only the first element in the list. One element (red) is removed from the input data and inserted in-place into the sorted list with each iteration.</p>
<img alt="" src="https://upload.wikimedia.org/wikipedia/commons/0/0f/Insertion-sort-example-300px.gif" style="height:180px; width:300px" />
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/04/sort1linked-list.jpg" style="width: 422px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [4,2,1,3]
<strong>Output:</strong> [1,2,3,4]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/04/sort2linked-list.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [-1,5,3,4,0]
<strong>Output:</strong> [-1,0,3,4,5]
</pre>

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

<ul>
	<li>The number of nodes in the list is in the range <code>[1, 5000]</code>.</li>
	<li><code>-5000 &lt;= Node.val &lt;= 5000</code></li>
</ul>

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def insertionSortList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy = ListNode()
        curr = head

        while curr:
            # At each iteration, we insert an element into the resulting list.
            prev = dummy

            # find the position to insert the current node
            while prev.next and prev.next.val <= curr.val:
                prev = prev.next

            next = curr.next
            # insert the current node to the new list
            curr.next = prev.next
            prev.next = curr

            # moving on to the next iteration
            curr = next

        return dummy.next        

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