You are given an array of k linked-lists lists, each linked-list is sorted in ascending order. Merge all the linked-lists into one sorted linked-list and return it.

Example 1:

Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
  1->4->5,
  1->3->4,
  2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6

Example 2:

Input: lists = []
Output: []

Example 3:

Input: lists = [[]]
Output: []

Constraints:

• k == lists.length
• 0 <= k <= 10^4
• 0 <= lists[i].length <= 500
• -10^4 <= lists[i][j] <= 10^4
• lists[i] is sorted in ascending order.
• The sum of lists[i].length won't exceed 10^4.

这道题有一个便捷条件，就是每个list都是sorted，所以我们只需要比较几个list的当前node，每次把最小的取出，加入到结果的list里面即可。有限数目下取最小值，这个可以用PriorityQueue来实现。先把所有的ListNode的头都加到PriorityQueue里，然后取出最小的加到结果queue里，再把这个取出node的下一个node加到PriorityQueue里。
注意点：

• PriorityQueue的格式写法，尤其是Comparator
• Comparator的法则
  • 结果为负数，第一个比第二个靠前
  • 结果为正数，第一个比第二个靠后
  • 结果为0，两个排序不分先后
• 加入PriorityQueue的时候要注意检查是不是null
• 我比较喜欢加入一个dummy node来帮助返回。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) {
            return null;
        }
        
        ListNode dummy = new ListNode(0);
        ListNode head = dummy;
        
        PriorityQueue<ListNode> pq = new PriorityQueue<ListNode>(lists.length, 
            new Comparator<ListNode>(){
                public int compare(ListNode o1, ListNode o2) {
                    return o1.val - o2.val;
                }
            });
        
        for (int i = 0; i < lists.length; i++) {
            if (lists[i] != null) {
                pq.add(lists[i]);
            }
        }
        
        while(!pq.isEmpty()) {
            ListNode tmpMin = pq.poll();
            if (tmpMin.next != null) {
               pq.add(tmpMin.next); 
            }
            head.next = tmpMin;
            head = tmpMin;
        }
        return dummy.next;
    }
    
}