LeetCode – Sort List (Java)

LeetCode - Sort List:

Sort a linked list in O(n log n) time using constant space complexity.

Keys for solving the problem

  1. Break the list to two in the middle
  2. Recursively sort the two sub lists
  3. Merge the two sub lists

This is my accepted answer for the problem.

package algorithm.sort;
 
class ListNode {
	int val;
	ListNode next;
 
	ListNode(int x) {
		val = x;
		next = null;
	}
}
 
public class SortLinkedList {
 
	// merge sort
	public static ListNode mergeSortList(ListNode head) {
 
		if (head == null || head.next == null)
			return head;
 
		// count total number of elements
		int count = 0;
		ListNode p = head;
		while (p != null) {
			count++;
			p = p.next;
		}
 
		// break up to two list
		int middle = count / 2;
 
		ListNode l = head, r = null;
		ListNode p2 = head;
		int countHalf = 0;
		while (p2 != null) {
			countHalf++;
			ListNode next = p2.next;
 
			if (countHalf == middle) {
				p2.next = null;
				r = next;
			}
			p2 = next;
		}
 
		// now we have two parts l and r, recursively sort them
		ListNode h1 = mergeSortList(l);
		ListNode h2 = mergeSortList(r);
 
		// merge together
		ListNode merged = merge(h1, h2);
 
		return merged;
	}
 
	public static ListNode merge(ListNode l, ListNode r) {
		ListNode p1 = l;
		ListNode p2 = r;
 
		ListNode fakeHead = new ListNode(100);
		ListNode pNew = fakeHead;
 
		while (p1 != null || p2 != null) {
 
			if (p1 == null) {
				pNew.next = new ListNode(p2.val);
				p2 = p2.next;
				pNew = pNew.next;
			} else if (p2 == null) {
				pNew.next = new ListNode(p1.val);
				p1 = p1.next;
				pNew = pNew.next;
			} else {
				if (p1.val < p2.val) {
					// if(fakeHead)
					pNew.next = new ListNode(p1.val);
					p1 = p1.next;
					pNew = pNew.next;
				} else if (p1.val == p2.val) {
					pNew.next = new ListNode(p1.val);
					pNew.next.next = new ListNode(p1.val);
					pNew = pNew.next.next;
					p1 = p1.next;
					p2 = p2.next;
 
				} else {
					pNew.next = new ListNode(p2.val);
					p2 = p2.next;
					pNew = pNew.next;
				}
			}
		}
 
		// printList(fakeHead.next);
		return fakeHead.next;
	}
 
	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 = mergeSortList(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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  • Saikiran Narlapuram
  • Arnaud Hebert

    Works fine if you change int count = 0; with int count = 1;

  • henrycan1414

    it is the same since p1.val == p2. val

  • Neel Sheyal

    If the list is empty or only one element in it, then return the list.
    Divide the linked list into two parts.
    Sort these two parts recursively.
    Merge the sorted parts.

    For explanation and code http://www.algoqueue.com/algoqueue/default/view/851968/merge-sort-on-linkedlist

  • Frank

    This line
    pNew.next.next = new ListNode(p1.val);

    Should be

    pNew.next.next = new ListNode(p2.val);

  • anchao1987

    maybe newHead is not nessary.

  • tia

    A little bit change to merge function works. Create extra nodes is unnecessary

    public ListNode merge(ListNode l, ListNode r){

    ListNode lp = l, rp = r;

    ListNode newhead = new ListNode(-1);

    ListNode cur = newhead;

    while(lp!=null || rp!=null){

    if(lp==null){

    cur.next = rp;

    break;

    }else if(rp==null){

    cur.next = lp;

    break;

    }else{

    if(lp.val <= rp.val){

    cur.next = lp;

    lp = lp.next;

    cur = cur.next;

    }else {

    cur.next = rp;

    rp = rp.next;

    cur = cur.next;

    }

    }

    }

    return newhead.next;

    }

  • Raghav

    I’m getting stack overflow error for size of length greater than 3.

  • Lister

    Definitely not correct. You are creating new nodes.

  • Wei Qiu

    It’s not correct if you consider the space used by call stack.