Sort List | Leetcode

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

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Solution: Probably the only way to do this using constant memory is to use merge sort. We use the technique of finding the middle of the linked list using a slow pointer and a fast pointer. So we can break down a linked lists into two parts, and then sort each part and merge them together.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null) return null;
        if (head.next == null) return head;
        
        // Find the middle of the linked list.
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        
        // Break up the linked list into two parts.
        ListNode first = head;
        ListNode second = slow.next;
        slow.next = null;
        
        // Sort each part separately.
        first = sortList(first);
        second = sortList(second);
        
        // Merge the two linked lists.
        ListNode curr = null;
        while (first != null || second != null) {
            ListNode node;
            if (first == null) {
                node = second;
                second = second.next;
            } else if (second == null) {
                node = first;
                first = first.next;
            } else if (first.val < second.val) {
                node = first;
                first = first.next;
            } else {
                node = second;
                second = second.next;
            }
            if (curr == null) {
                head = curr = node;
            } else {
                curr.next = node;
                curr = node;
            }
        }
        
        return head;
    }
}