Merge k Sorted Lists-LeetCode#23

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

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

合并k排序列表
合并k个已排序的链表并将其作为一个排序列表返回。分析并描述其复杂性。
第一时间想到的就是循环递归合并每个链表，直接上代码:

public class MergeKSortedLists {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists.length == 0) {
            return null;
        }else if (lists.length == 1){
            return lists[0];
        }
        ListNode res = lists[0];
        for (int i = 0; i < lists.length - 1; i++) {
            res = dfs(res, lists[i + 1]);
        }
        return res;
    }
    public ListNode dfs(ListNode l1, ListNode l2) {
        if (l1 == null){
            return l2;
        }
        if (l2 == null){
            return l1;
        }
        if (l1.val < l2.val) {
            l1.next = dfs(l1.next, l2);
            return l1;
        } else {
            l2.next = dfs(l1, l2.next);
            return l2;
        }
    }
    public static void main(String[] args) {
        MergeKSortedLists mergeKSortedLists = new MergeKSortedLists();
        ListNode l1 = new ListNode(1);
        l1.next = new ListNode(4);
        l1.next.next = new ListNode(5);
        ListNode l2 = new ListNode(1);
        l2.next = new ListNode(3);
        l2.next.next = new ListNode(4);
        ListNode l3 = new ListNode(2);
        l3.next = new ListNode(6);
        ListNode[] listNodes = {l1,l2,l3};
        mergeKSortedLists.mergeKLists(listNodes);
    }
}
class ListNode {
 int val;
 ListNode next;
 ListNode(int x) {
 val = x;
 }
}

看了 vote 最高的 ac 算法，真是大开眼界。

public class Solution {
    public ListNode mergeKLists(List<ListNode> lists) {
        if (lists==null||lists.size()==0) return null;
        PriorityQueue<ListNode> queue= new PriorityQueue<ListNode>(lists.size(),new Comparator<ListNode>(){
            @Override
            public int compare(ListNode o1,ListNode o2){
                if (o1.val<o2.val)
                    return -1;
                else if (o1.val==o2.val)
                    return 0;
                else
                    return 1;
            }
        });
        ListNode dummy = new ListNode(0);
        ListNode tail=dummy;
        for (ListNode node:lists)
            if (node!=null)
                queue.add(node);
        while (!queue.isEmpty()){
            tail.next=queue.poll();
            tail=tail.next;
            if (tail.next!=null)
                queue.add(tail.next);
        }
        return dummy.next;
    }
}

先熟悉一下 PriorityQueue 的数据结构