Insertion Sort List LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
Share:

Insertion Sort 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

Insertion Sort Solutions

Time: O(n)
Space: O(1)

C++

 class Solution {
 public:
  ListNode* insertionSortList(ListNode* head) {
    ListNode dummy(0);
    ListNode* prev = &dummy;  // the last (largest) of the sorted list

    while (head) {                  // current inserting node
      ListNode* next = head->next;  // cache next inserting node
      if (prev->val >= head->val)   // `prev` >= current inserting node
        prev = &dummy;              // move `prev` to the front
      while (prev->next && prev->next->val < head->val)
        prev = prev->next;
      head->next = prev->next;
      prev->next = head;
      head = next;  // update current inserting node
    }

    return dummy.next;
  }
};

Java

 class Solution {
  public ListNode insertionSortList(ListNode head) {
    ListNode dummy = new ListNode(0);
    ListNode prev = dummy; // the last (largest) of the sorted list

    while (head != null) {       // current inserting node
      ListNode next = head.next; // cache next inserting node
      if (prev.val >= head.val)  // `prev` >= current inserting node
        prev = dummy;            // move `prev` to the front
      while (prev.next != null && prev.next.val < head.val)
        prev = prev.next;
      head.next = prev.next;
      prev.next = head;
      head = next; // update current inserting node
    }

    return dummy.next;
  }
}

Python

class Solution:
  def insertionSortList(self, head: ListNode) -> ListNode:
    dummy = ListNode(0)
    curr = head

    while curr:
      prev = dummy
      while prev.next and prev.next.val < curr.val:
        prev = prev.next
      next = curr.next
      curr.next = prev.next
      prev.next = curr
      curr = next

    return dummy.next

Watch Tutorial

Checkout more Solutions here

Leave a Comment

Your email address will not be published. Required fields are marked *

x