23. Merge k Sorted List

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Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

解法1： PriorityQueue, O(NlogM)

M是queue的长度，N是所有node的总数。
一个简单的PriorityQueue的应用。

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) {
            return null;
        }
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        Queue<ListNode> queue = new PriorityQueue<>((a,b) -> a.val - b.val);
        for (int i = 0; i < lists.length; i++) {
            if (lists[i] != null) {
                queue.offer(lists[i]);            
            }
        }
        while (!queue.isEmpty()) {
            ListNode next = queue.poll();
            tail.next = next;
            tail = tail.next;
            // add another node if exist
            if (next.next != null) {
                queue.offer(next.next);
            }
        }
        return dummy.next;
    }
}