Sort List

Problem

Given the head of a linked list, return the list after sorting it in ascending order.

 

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]

Example 3:

Input: head = []
Output: []

 

Constraints:

• The number of nodes in the list is in the range [0, 5 * 104].
• -105 <= Node.val <= 105

Solution

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var sortList = function(head) {
    // list size 0 or 1
    if (!head?.next) {
        return head;
    }

    // split list into two
    let slow = head;
    let fast = head;
    while (fast.next?.next) {
        slow = slow.next;
        fast = fast.next?.next;
    }
    let l1 = head;
    let l2 = slow.next;
    slow.next = null;

    // divide and conquer
    l1 = sortList(l1);
    l2 = sortList(l2);

    return merge(l1, l2);
};

var merge = function(l1, l2) {
    const dummy = new ListNode();
    let cur = dummy;
    while (l1 && l2) {
        if (l1.val < l2.val) {
            cur.next = l1;
            l1 = l1.next;
        } else {
            cur.next = l2;
            l2 = l2.next;
        }
        cur = cur.next;
    }
    cur.next = l1 ?? l2;
    return dummy.next;
};

We will implement a classic merge sort algorithm to sort the linked list head.

Follow-up

Can you sort the linked list in O(n logn) time and O(1) memory (i.e. constant space)?

Solution

This is non-trivial, but is doable if we implement a bottom-up version of merge sort instead of top-down. This way, we no longer use recursion (ie. get rid of the additional space used to store the call stack).