问题

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。

思路

public class ListNode {
    private int val;
    private ListNode next;
​
    public ListNode mergeKLists(ListNode[] lists) {
        Queue<Node> queue = new PriorityQueue<Node>();
        for (ListNode list : lists) {
            if (list != null) {
                queue.add(new Node(list.val, list));
            }
        }
        ListNode head = new ListNode();
        ListNode tail = head;
        while (!queue.isEmpty()) {
            Node min = queue.poll();
            tail.next = min.node;
            tail = tail.next;
​
            if (min.node.next != null) {
                queue.add(new Node(min.node.next.val, min.node.next));
            }
        }
        return head.next;
    }
​
    class Node implements Comparable<Node> {
        private int val;
        private ListNode node;
​
        public Node(int val, ListNode node) {
            this.val = val;
            this.node = node;
        }
​
        public int compareTo(Node o) {
            return this.val - o.val;
        }
    }
}

update 20210317

    public ListNode mergeKLists(ListNode[] lists) {
        Queue<ListNode> queue = new PriorityQueue<>(new Comparator<ListNode>() {
            @Override
            public int compare(ListNode o1, ListNode o2) {
                return o1.val - o2.val;
            }
        });

        for (ListNode node : lists) {
            if (node != null) {
                queue.add(node);
            }
        }

        ListNode head = null;
        ListNode p = null;
        while (!queue.isEmpty()) {
            ListNode node = queue.poll();
            if (head == null) {
                head = node;
            } else {
                p.next = node;
            }
            p = node;
            if (node.next != null) {
                queue.add(node.next);
            }
        }

        return head;
    }

update20220412

public ListNode mergeKLists(ListNode[] lists) {
    if (lists == null || lists.length == 0) {
        return null;
    }
    ListNode dummyHead = new ListNode();
    ListNode p = dummyHead;
    Queue<ListNode> queue = new PriorityQueue<>((n1, n2) -> n1.val - n2.val);
    for (ListNode list : lists) {
        if (list != null) {
            queue.offer(list);
        }
    }
    while (!queue.isEmpty()) {
        ListNode temp = queue.poll();
        p.next = temp;
        p = temp;
        if (temp.next != null) {
            queue.offer(temp.next);
        }
    }
    return dummyHead.next;
}

update20230419

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        ListNode head = new ListNode(0, null);
        int size = lists.length;
        Queue<ListNode> queue = new PriorityQueue<ListNode>((o1, o2) -> o1.val - o2.val);
        for (int i = 0; i < size; i++) {
            if (lists[i] != null) {
                queue.offer(lists[i]);
            }  
        }
        ListNode p = head;
        while (!queue.isEmpty()) {
            ListNode node = queue.poll();
            if (node.next != null) {
                queue.offer(node.next);
            }
            p.next = node;
            p = p.next;
        }
        return head.next;
    }
}

复杂度

时间复杂度：k*n*logk。k是指链表数组的个数。n是指每个链表的长度k*n表示每个节点都要被遍历一次。logk表示优先队列增加、删除节点后调整的时间复杂度。

空间复杂度：k。优先队列中元素的个数。

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