147. Insertion Sort List

Explanation:

Algorithmic Idea:
1. We start with an empty sorted list and iterate through the input list.
2. For each element in the input list, we remove it and insert it in the correct position in the sorted list.
3. To insert an element into the sorted list, we compare it with each element from the beginning of the sorted list and find the correct position to insert it.
Step-by-Step Iterations:
- For input list [4,2,1,3]:
  1. Start with an empty sorted list.
  2. Insert 4 into the sorted list: [4]
  3. Insert 2 into the sorted list: [2,4]
  4. Insert 1 into the sorted list: [1,2,4]
  5. Insert 3 into the sorted list: [1,2,3,4]
Time Complexity: O(n^2) where n is the number of nodes in the linked list.
Space Complexity: O(1)

:

class Solution {
    public ListNode insertionSortList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode dummy = new ListNode(-1);
        ListNode curr = head;
        
        while (curr != null) {
            ListNode prev = dummy;
            ListNode nextNode = curr.next;
            
            while (prev.next != null && prev.next.val < curr.val) {
                prev = prev.next;
            }
            
            curr.next = prev.next;
            prev.next = curr;
            curr = nextNode;
        }
        
        return dummy.next;
    }
}

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Explanation:

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