147. Insertion Sort List

Problem Statement

link: https://leetcode.com/problems/insertion-sort-list/ https://leetcode.cn/problems/insertion-sort-list/

Given the head of a singly linked list, sort the list using insertion sort, and return the sorted list’s head.

The steps of the insertion sort algorithm:

Insertion sort iterates, consuming one input element each repetition and growing a sorted output list. 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. It repeats until no input elements remain. 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.

Example:

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

Input: head = [-1,5,3,4,0] Output: [-1,0,3,4,5]

Solution Approach

The solution uses the classic insertion sort technique, but applies it on a singly linked list, where elements are moved by changing the pointers instead of swapping.

Algorithm

1. Create a new dummy node to act as the starting point of the sorted list.
2. Traverse the original list, for each node, find its position in the sorted list and insert it there by adjusting the pointers.
3. Repeat the process until the original list is exhausted and return the next pointer of the dummy node, which will be the head of the sorted list.

Implement

    class Solution:
    def insertionSortList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        pre = dummy = ListNode()
        node = head
        while node:
            cur = node
            node = node.next
            if cur.val < pre.val:
                pre = dummy
            while pre.next and cur.val > pre.next.val:
                pre = pre.next
            cur.next = pre.next
            pre.next = cur
        return dummy.next
 

    
    

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