【每日一题】【小根堆&边出队边入队后续节点&注意判空】23. 合并K个升序链表-211128/220213

给你一个链表数组，每个链表都已经按升序排列。

请你将所有链表合并到一个升序链表中，返回合并后的链表。

答案1（参数是数组）：

/**
 * 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.length == 0) {
            return null;
        }
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        //后大前小
        PriorityQueue<ListNode> pq = new PriorityQueue<>((o1, o2) -> o1.val - o2.val);
        for(ListNode list : lists) {
            if(list == null) {
                continue;
            }
            pq.add(list);
        }
        while(!pq.isEmpty()) {
            ListNode next = pq.poll();
            cur.next = next;
            cur = cur.next;
            if(next.next != null) {
                pq.add(next.next);
            }
        }
        return dummy.next;
    }
}

答案2：（参数是ArrayList）

import java.util.*;
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode mergeKLists(ArrayList<ListNode> lists) {
        ListNode res = new ListNode(0);
        ListNode cur = res;
        PriorityQueue<ListNode> queue =
            new PriorityQueue<>((o1, o2) -> o1.val - o2.val);
        for(ListNode list : lists) {
            if(list != null) {
                queue.add(list);
            }
        }
        while(!queue.isEmpty()) {
            ListNode node = queue.poll();
            if(node == null) {
                continue;
            }
            cur.next = node;
            cur = cur.next;
            if(node.next != null) {
                queue.offer(node.next);
            }
        }
        return res.next;
    }
}

自己方法：（超时）

import java.util.*;
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode mergeKLists(ArrayList<ListNode> lists) {
        //使用优先队列（小根堆）
        PriorityQueue<ListNode> queue =
            new PriorityQueue<>((o1, o2) -> o1.val - o2.val);
        ListNode res = new ListNode(0);
        ListNode cur = res;
        Iterator<ListNode> iter = lists.iterator();
        while(iter.hasNext()) {
            ListNode list = iter.next();
            while(list != null) {
                queue.add(list);
                list = list.next;
            }
        }
        while(!queue.isEmpty()) {
            cur.next = queue.poll();
            cur = cur.next;
        }
        return res.next;
    }
}