LeetCode Solution – Sort a linked list using insertion sort in Java

Category: Algorithms November 19, 2012

Insertion Sort List:

Sort a linked list using insertion sort.

This is my accepted answer for LeetCode problem - Sort a linked list using insertion sort in Java. It is a complete program.

Before coding for that, here is an example of insertion sort from wiki. You can get an idea of what is insertion sort.

insertion-sort-example

Code:

package algorithm.sort;
 
class ListNode {
	int val;
	ListNode next;
 
	ListNode(int x) {
		val = x;
		next = null;
	}
}
 
public class SortLinkedList {
	public static ListNode insertionSortList(ListNode head) {
 
		if (head == null || head.next == null)
			return head;
 
		ListNode newHead = new ListNode(head.val);
		ListNode pointer = head.next;
 
		// loop through each element in the list
		while (pointer != null) {
			// insert this element to the new list
 
			ListNode innerPointer = newHead;
			ListNode next = pointer.next;
 
			if (pointer.val <= newHead.val) {
				ListNode oldHead = newHead;
				newHead = pointer;
				newHead.next = oldHead;
			} else {
				while (innerPointer.next != null) {
 
					if (pointer.val > innerPointer.val && pointer.val <= innerPointer.next.val) {
						ListNode oldNext = innerPointer.next;
						innerPointer.next = pointer;
						pointer.next = oldNext;
					}
 
					innerPointer = innerPointer.next;
				}
 
				if (innerPointer.next == null && pointer.val > innerPointer.val) {
					innerPointer.next = pointer;
					pointer.next = null;
				}
			}
 
			// finally
			pointer = next;
		}
 
		return newHead;
	}
 
	public static void main(String[] args) {
		ListNode n1 = new ListNode(2);
		ListNode n2 = new ListNode(3);
		ListNode n3 = new ListNode(4);
 
		ListNode n4 = new ListNode(3);
		ListNode n5 = new ListNode(4);
		ListNode n6 = new ListNode(5);
 
		n1.next = n2;
		n2.next = n3;
		n3.next = n4;
		n4.next = n5;
		n5.next = n6;
 
		n1 = insertionSortList(n1);
 
		printList(n1);
 
	}
 
	public static void printList(ListNode x) {
		if(x != null){
			System.out.print(x.val + " ");
			while (x.next != null) {
				System.out.print(x.next.val + " ");
				x = x.next;
			}
			System.out.println();
		}
 
	}
}

Output:

2 3 3 4 4 5

Category >> Algorithms

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Atul

I don’t think so you need this if condition, without this if condition also u r code works.

if (pointer.val <= newHead.val) {
ListNode oldHead = newHead;
newHead = pointer;
newHead.next = oldHead;
}
gao can

How about my solution: without extra space
Rupam Ranjan Rana

How to do it High to low?
junmin

I believe you can do insertion sort in place when using linked list
Zack

hi dude, insertion sort has O(n^2) time complexity.

LeetCode Solution – Sort a linked list using insertion sort in Java

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