148. Sort List

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
/*
Refer: https://leetcode.com/problems/sort-list/discuss/46712/Bottom-to-up(not-recurring)-with-o(1)-space-complextity-and-o(nlgn)-time-complextity

Time Complexity: O(n logn)
Outer for loop runs for log n times. Inner while loop runs for 2 * n times.

Space Complexity: O(1)
This algorithm uses constant space.
*/
class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        ListNode cur = head;
        int length = 0;
        while (cur != null) {
            length++;
            cur = cur.next;
        }

        ListNode dummyNode = new ListNode(Integer.MIN_VALUE);
        dummyNode.next = head;
        for (int step = 1; step < length; step <<= 1) {
            cur = dummyNode.next;
            ListNode tail = dummyNode;
            while (cur != null) {
                ListNode left = cur;
                ListNode right = split(left, step);
                cur = split(right, step);
                tail = merge(tail, left, right);
            }
        }

        return dummyNode.next;
    }

    public ListNode split(ListNode head, int n) {
        for (int i = 1; i < n && head != null; i++) {
            head = head.next;
        }
        if (head == null) {
            return null;
        }
        ListNode next = head.next;
        head.next = null;
        return next;
    }

    public ListNode merge(ListNode head, ListNode l1, ListNode l2) {
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                head.next = l1;
                l1 = l1.next;
            } else {
                head.next = l2;
                l2 = l2.next;
            }
            head = head.next;
        }
        head.next = l1 == null ? l2 : l1;
        while (head.next != null) {
            head = head.next;
        }
        return head;
    }
}