LeetCode – Sort List (Java)

LeetCode - Sort List:

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

1. Analysis

If this problem does not have the constant space limitation, we can easily sort using a sorting method from Java SDK. With the constant space limitation, we need to do some pointer manipulation.

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

2. Acceptable Java Solution

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

3. Simplified Java Solution

When I revisit this problem in 2018, I wrote it the following way which is more concise.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
        if(head==null || head.next==null){
            return head;
        }
 
        //partition the list
        ListNode p1 = head;
        ListNode firstEnd = getFirstEnd(head);
        ListNode p2 = firstEnd.next;
        firstEnd.next = null;
 
        //sort each list
        p1 = sortList(p1);
        p2 = sortList(p2);
 
        //merge two lists
        return merge(p1, p2);
    }
 
    //get the list partition point
    private ListNode getFirstEnd(ListNode head){
        ListNode p1 = head;
        ListNode p2 = head;
        while(p1!=null && p2!=null){
            if(p2.next==null||p2.next.next==null){
                return p1;
            }
 
            p1 = p1.next;
            p2 = p2.next.next;
        }
 
        return head;
    }
 
    //merge two list
    private ListNode merge(ListNode n1, ListNode n2){
        ListNode head = new ListNode(0);
        ListNode p = head;
        ListNode p1 = n1;
        ListNode p2 = n2; 
        while(p1!=null && p2!=null){
            if(p1.val<p2.val){
                p.next = p1;
                p1 = p1.next;
            }else{
                p.next = p2;
                p2 = p2.next;
            }
 
            p = p.next;
        }
 
        if(p1!=null){
            p.next = p1;
        }
 
        if(p2!=null){
            p.next = p2;
        }
 
        return head.next;
    }
}
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.