0147. Insertion Sort List

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

Description

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:

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.

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 1:

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

Example 2:

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

Constraints:

• The number of nodes in the list is in the range [1, 5000].

• -5000 <= Node.val <= 5000

ac

know insertion sort

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode insertionSortList(ListNode head) {
        // prev/head, walk, if head>prev continue, else stop.
        // extract head -> tmp
        // insert tmp in sorted part, t1/t2, if t2>head stop, insert.
        // loop, compare.

        // corner case
        if (head == null || head.next == null) return head;

        // walk
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode prev = head;
        head = head.next;

        while (head != null) {
            if (head.val >= prev.val) {
                prev = head;
                head = head.next;
                continue;
            }
            // stop, extract
            ListNode tmp = head;
            head = head.next;
            prev.next = head;
            // find insertion point in sorted part
            ListNode t1 = dummy, t2 = t1.next;
            while (t2.val < tmp.val) {
                t1 = t2;
                t2 = t2.next;
            }
            // insert tmp
            tmp.next = t2;
            t1.next = tmp;
        }

        return dummy.next;
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode insertionSortList(ListNode head) {
        ListNode dummy = new ListNode(0);
        ListNode prev = dummy;

        while (head != null) {
            ListNode tmp = head.next; // for further usage
            if (head.val < prev.val) prev = dummy;  // reset to head, find insertion point
            while (prev.next != null && prev.next.val < head.val) 
                prev = prev.next;

            head.next = prev.next;
            prev.next = head;
            head = tmp;
        }

        return dummy.next;
    }
}
/*
dummy head as sorted list, when old list head < new list, find insertion point
*/