23Merge k Sorted Lists

题目链接：https://leetcode.com/problems/merge-k-sorted-lists/

题目：

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

解题思路：
采用的数据结构是优先级队列。
将参加归并的每个链表的头结点都入队列，这样每次可以从队列中取出一个最小的结点放入新链表中，如果选中的结点的后继不为空，就让其后继入队列。

以前基本没怎么用过优先级队列，小小记录下。
优先级队列 PriorityQueue < T > 方法小结：
boolean add(E e)
Inserts the specified element into this priority queue.
E peek()
Retrieves, but does not remove, the head of this queue, or returns null if this queue is empty.
E poll()
Retrieves and removes the head of this queue, or returns null if this queue is empty.

优先级队列比较结点时采用 Comparator 接口的 compare(T o1, T o2) 方法实现。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode mergeKLists(ListNode[] lists) {        if(lists == null || lists.length == 0)            return null;        PriorityQueue<ListNode> queue = new PriorityQueue(lists.length, new Comparator<ListNode>() {                @Override                public int compare(ListNode l1, ListNode l2) {                    return l1.val - l2.val;                }        });        for(int i = 0; i < lists.length; i ++) {            if(lists[i] != null)                queue.add(lists[i]);        }        ListNode res = new ListNode(0);        ListNode p = res;        while(!queue.isEmpty()) {            ListNode tmp = queue.poll();            p.next = tmp;            p = p.next;            if(tmp.next != null)                queue.add(tmp.next);        }        return res.next;    }}

130 / 130 test cases passed.Status: AcceptedRuntime: 424 ms