题解 | #合并k个已排序的链表#

用堆来存储每个链表的当前头结点

    public ListNode mergeKLists(ArrayList<ListNode> lists) {
        ListNode headNode = new ListNode(-1);
        ListNode r = headNode;
        PriorityQueue<ListNode> heap = new PriorityQueue<>((a, b) -> a.val - b.val);
        for(int i = 0; i < lists.size(); i++){
            ListNode node = lists.get(i);
            if(node != null){
                heap.offer(node);
            }
        }
        while(!heap.isEmpty()){
            ListNode p = heap.poll();
            r.next = p;
            r = p;
            if(p.next != null){
                p = p.next;
                heap.offer(p);
            }
        }
        return headNode.next;
    }
}