Question

Formatted question description: https://leetcode.ca/all/147.html

147	Insertion Sort List

Sort a linked list using insertion sort.
https://upload.wikimedia.org/wikipedia/commons/0/0f/Insertion-sort-example-300px.gif

Algorithm of Insertion Sort:
    1. Insertion sort iterates, consuming one input element each repetition,
        and growing a sorted output list.
    2. 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.
    3. It repeats until no input elements remain.

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5


@tag-linkedlist

Algorithm

One by one element is taken out of the original linked list and then inserted into the new linked list in order.

The time complexity is O(n^2), which is an algorithm that is not very efficient, but the space complexity is O(1), in exchange for high time complexity for low space complexity.

Code

Java

public class Insertion_Sort_List {

	/**
	 * Definition for singly-linked list.
	 * public class ListNode {
	 *     int val;
	 *     ListNode next;
	 *     ListNode(int x) { val = x; }
	 * }
	 */
	public class Solution {

	    public ListNode insertionSortList(ListNode head) {

	        if (head == null) {
	            return null;
	        }

	        // for current tail, maintain its prev and itself, for moving around
	        ListNode dummy = new ListNode(0);
	        // dummy.next = head; // @note:@memorize: here is the start of wrong

	        /*
	            @note: I think the key is to cut the list into 2 half, un-linked
	                    I was trying to move node around using i, j to indicate the progress of sorted part, like insert-sort in arrays
	                    but it was nightmare moving pointers

	                    the key is the same as the below solution which create new nodes
	                    but here no need to create new nodes, just pointer moving
	                    and key is the same, keep 2 separate, un-linked, un-related lists
	        */

            ListNode current = head;
	        while (current != null) {
                // every time, scanner will scan from begging of sorted list, to find where to insert
	            ListNode scanner = dummy;

	            // @note:@memorize: here use scanner.next, so that insert postion is at scanner next
	            while (scanner.next != null && scanner.next.val < current.val) {
	                scanner = scanner.next;
	            }

	            // record for next loop
	            ListNode nextRoundStart = current.next;

	            // insert
                current.next = scanner.next;
	            scanner.next = current;

	            // update
                current = nextRoundStart;

	        }

	        return dummy.next;
	    }

	}

    // create a separate list, extra space...
	public class Solution_extra_space {
	      public ListNode insertionSortList(ListNode head) {
	          if (head == null || head.next == null)  return head;

	          ListNode dummy = new ListNode(0);
	          dummy.next = new ListNode(head.val);

	          ListNode current = head.next; // add head before loop

	          while(current != null) {

	              ListNode prev = dummy;
	              ListNode newnode = new ListNode(current.val);

	              while(prev.next != null && newnode.val > prev.next.val) {
	                  prev = prev.next;
	              }

	              // insert
	              ListNode oldPrevNext = prev.next;
	              prev.next = newnode;
	              newnode.next = oldPrevNext;

	              current = current.next; // loop through the list
	          }


	          return dummy.next;
	      }
	  }

}

// OJ: https://leetcode.com/problems/insertion-sort-list/
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
    ListNode* insertionSortList(ListNode* head) {
        ListNode dummy;
        while (head) {
            auto node = head;
            head = head->next;
            auto p = &dummy;
            while (p->next && p->next->val < node->val) p = p->next;
            node->next = p->next;
            p->next = node;
        }
        return dummy.next;
    }
};

# 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: ListNode) -> ListNode:
        if head is None or head.next is None:
            return head
        dummy = ListNode(head.val, head)
        pre, cur = dummy, head
        while cur:
            if pre.val <= cur.val:
                pre, cur = cur, cur.next
                continue
            p = dummy
            while p.next.val <= cur.val:
                p = p.next
            t = cur.next
            cur.next = p.next
            p.next = cur
            pre.next = t
            cur = t
        return dummy.next

############

class Solution(object):
  def insertionSortList(self, head):
    p = dummy = ListNode(0)
    cur = dummy.next = head
    while cur and cur.next:
      val = cur.next.val
      if cur.val < val:
        cur = cur.next
        continue
      if p.next.val > val:
        p = dummy
      while p.next.val < val:
        p = p.next
      new = cur.next
      cur.next = new.next
      new.next = p.next
      p.next = new
    return dummy.next

147 - Insertion Sort List

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