23. Merge k Sorted Lists

// LeetCode URL: https://leetcode.com/problems/add-two-numbers/
import java.util.PriorityQueue;

//   Definition for singly-linked list.
class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {
        val = x;
    }
}

/**
 * Using Priority Queue
 *
 * Time Complexity: O(N log K)
 *
 * Space Complexity: O(K)
 *
 * N = Total number of all nodes in all lists. K = Number of lists.
 */
class Solution1 {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) {
            return null;
        }
        if (lists.length == 1) {
            return lists[0];
        }

        PriorityQueue<ListNode> queue = new PriorityQueue<>(lists.length, (a, b) -> (a.val - b.val));

        ListNode dummyNode = new ListNode(-1);
        ListNode prevNode = dummyNode;

        for (ListNode listHead : lists) {
            if (listHead != null) {
                queue.offer(listHead);
            }
        }

        while (!queue.isEmpty()) {
            prevNode.next = queue.poll();
            prevNode = prevNode.next;

            if (prevNode.next != null) {
                queue.offer(prevNode.next);
            }
        }

        return dummyNode.next;
    }
}

/**
 * Merge with Divide & Conquer
 *
 * Time Complexity: Divide and Merge operation forms a tree with height log K.
 * Suppose there are N total nodes in all the list.. Thus the total time
 * complexity is O(N * log K)
 *
 * Space Complexity: O(1)
 *
 * N = Total number of all nodes in all lists. K = Number of lists.
 */
class Solution2 {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) {
            return null;
        }
        if (lists.length == 1) {
            return lists[0];
        }

        int interval = 1; // Interval to find the index of lists to be merged.
        while (interval < lists.length) {
            for (int i = 0; i < lists.length - interval; i += interval * 2) {
                lists[i] = merge2Lists(lists[i], lists[i + interval]);
            }
            interval *= 2;
        }

        return lists[0];
    }

    private ListNode merge2Lists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        }
        if (l2 == null) {
            return l1;
        }

        ListNode dummyHead = new ListNode(-1);
        ListNode prevNode = dummyHead;
        ListNode p = l1;
        ListNode q = l2;

        while (p != null && q != null) {
            if (p.val <= q.val) {
                prevNode.next = p;
                p = p.next;
            } else {
                prevNode.next = q;
                q = q.next;
            }
            prevNode = prevNode.next;
            prevNode.next = null;
        }

        prevNode.next = p == null ? q : p;
        return dummyHead.next;
    }
}