Java算法 -- 单链表的反转单链表实现栈和队列以及双端队列K 个一组翻转链表

Posted 2022-08-18 CodeJiao

1. 单链表的反转

单链表的反转即把原来的单链表的指针顺序倒置。如下图所示：

需要注意的点是，Java中是值（传递引用也可以认为是特殊的值传递）传递，函数在改变完值后记得赋值给原来的值，变化才会生效。

详细请看这篇文章（总的来说就是如果不改变传递进来的引用副本的值，那么更改引用相关的属性是可以生效的。如果改变了引用副本的值，那么对原引用的相关属性的更改是不生效的。）

示例代码：

    static class Node 
        public int value;
        public Node next;

        public Node(int value) 
            this.value = value;
        

    

    public static Node test(Node head) 
        head = head.next;
        return head;
    

    public static void main(String[] args) 
        // 1. 构造一个单链表
        Node head = new Node(1);
        head.next = new Node(2);

        // 2. 调用test函数
        test(head);
        System.out.println(head.value);

        // 3. 调用test函数并赋值给head
        head = test(head);
        System.out.println(head.value);
    

运行结果：

下面我们就可以正式的编写反转单链表的代码了：

    static class Node 
        public int value;
        public Node next;

        public Node(int value) 
            this.value = value;
        

    

    public static void main(String[] args) 
        // 1. 构造一个单链表
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);

        head = reverseSingleList(head);
        while (head != null) 
            System.out.print(head.value + " ");
            head = head.next;
        
    

    /**
     * 反转单链表
     *
     * @param head head是头结点
     * @return 反转后的单链表的头结点
     */
    public static Node reverseSingleList(Node head) 
        Node pre = null, next = null;
        while (head != null) 
            // next是为了可以往后继续遍历
            next = head.next;
            // head的next指向pre
            head.next = pre;
            // pre指向head
            pre = head;
            // head指向next
            head = next;
        
        return pre;
    

运行结果：

下面我们来解释原理：

补充：单链表介绍

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2. 双链表的反转

双链表的反转即把原来的双链表的指针顺序倒置。如下图所示：

编写双链表的反转的代码：

public class DoubleLinkedList 
    public static class DoubleNode 
        public char value;
        // next 指向下一个元素，last指向前一个元素
        public DoubleNode next, last;

        public DoubleNode(char value) 
            this.value = value;
        
    

    /**
     * 反转双向链表
     *
     * @param head 双向链表的头结点
     * @return 反转后双向链表的头结点
     */
    public static DoubleNode reverseDoubleList(DoubleNode head) 
        DoubleNode pre = null, next = null;
        while (head != null) 
            // next是为了可以往后继续遍历
            next = head.next;
            // head的next指向pre
            head.next = pre;
            // head的last指向next
            head.last = next;
            // pre指向head
            pre = head;
            // head指向next
            head = next;
        
        return pre;
    
    
	// 测试代码
    public static void main(String[] args) 
        DoubleNode node1 = new DoubleNode('a');
        DoubleNode node2 = new DoubleNode('b');
        DoubleNode node3 = new DoubleNode('c');

        node1.next = node2;
        node2.next = node3;
        node2.last = node1;
        node3.last = node2;

        node1 = reverseDoubleList(node1);
        while (node1 != null) 
            System.out.print(node1.value + " ");
            node1 = node1.next;
        
    

运行结果：

原理：

补充：双向链表介绍

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3. 单链表实现栈

栈介绍：

单链表实现栈：

public class MyStack<V> 
    private Node<V> head;
    private int size;

    public MyStack(int size) 
        this.size = size;
    

    public boolean isEmpty() 
        return size == 0;
    

    public int size() 
        return size;
    

    // 入栈
    public void push(V value) 
        Node<V> currentNode = new Node<>(value);
        if (head != null) 
            currentNode.next = head;
        
        head = currentNode;
        size++;
    

    // 出栈
    public V pop() 
        if (head != null) 
            V result = head.value;
            head = head.next;
            size--;
            return result;
        
        return null;
    
    // 查看栈顶元素（不弹出）

    public V peek() 
        return head == null ? null : head.value;
    

    static class Node<V> 
        public V value;
        public Node<V> next;

        public Node(V value) 
            this.value = value;
        
    

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4. 单链表实现队列

队列介绍:

单链表实现队列：

public class MyQueue<V> 
    // 指向队列头
    private Node<V> head;
    // 指向队列尾
    private Node<V> tail;
    // 队列大小
    private int size;

    public MyQueue(int size) 
        this.size = size;
    

    public boolean isEmpty() 
        return size == 0;
    

    public int size() 
        return size;
    

    // 网队列尾部插入一个元素
    public void offer(V value) 
        Node<V> currentNode = new Node<>(value);
        if (tail == null) 
            head = currentNode;
         else 
            // 让链表链接下去 保证结点可达
            tail.next = currentNode;
        
        // 让队列的尾部指针指向新结点
        tail = currentNode;
    

    // 弹出当前队列头的元素
    public V poll() 
        V result = null;
        if (head != null) 
            result = head.value;
            head = head.next;
            size--;
         else 
            tail = null;
        
        return result;
    

    // 获取当前队列头的元素（不弹出）
    public V peek() 
        if (head != null) 
            return head.value;
        
        return null;
    


class Node<V> 
    public V value;
    public Node<V> next;

    public Node(V value) 
        this.value = value;
    

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5. 用双向链表实现双端队列

双端队列相对于普通队列，双端队列的入队和出队操作在两端都可进行。

public class MYDoubleEndedQueue<V> 
    // 双端队列头
    private DoubleNode<V> head;
    // 双端队列尾
    private DoubleNode<V> tail;
    private int size;

    public MYDoubleEndedQueue(int size) 
        this.size = size;
    

    public boolean isEmpty() 
        return size == 0;
    

    public int size() 
        return size;
    

    // 头部插入元素
    public void pushHead(V value) 
        DoubleNode<V> currentNode = new DoubleNode<>(value);
        if (head == null) 
            tail = currentNode;
         else 
            currentNode.next = head;
            head.last = currentNode;
        
        head = currentNode;
        size++;
    

    // 尾部插入元素
    public void pushTail(V value) 
        DoubleNode<V> currentNode = new DoubleNode<>(value);
        if (head == null) 
            head = currentNode;
         else 
            tail.next = currentNode;
            currentNode.last = tail;
        
        tail = currentNode;
        size++;
    

    // 头部取出元素
    public V pollHead() 
        if (head == null) 
            return null;
        
        V result = head.value;
        if (head == tail) 
            head = null;
            tail = null;
         else 
            head = head.next;
            head.last = null;
        
        size--;
        return result;
    

    // 尾部取出元素
    public V pollTail() 
        if (head == null) 
            return null;
        
        V result = tail.value;
        if (head == tail) 
            head = null;
            tail = null;
         else 
            tail = tail.next;
            tail.next = null;
        
        size--;
        return result;
    


class DoubleNode<V> 
    public V value;
    public DoubleNode<V> last;
    public DoubleNode<V> next;

    public DoubleNode(V value) 
        this.value = value;
    

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6. K 个一组翻转链表

题目要求：力扣题目链接

代码实现：

public class ReverseNodesInKGroup 
    // 不要提交这个类
    public static class ListNode 
        public int val;
        public ListNode next;
    

    public static ListNode reverseKGroup(ListNode head, int k) 
        ListNode start = head;
        ListNode end = getKGroupEnd(start, k);
        if (end == null) 
            return head;
        
        // 第一组凑齐了！
        head = end;
        reverse(start, end);
        // 上一组的结尾节点
        ListNode lastEnd = start;
        while (lastEnd.next != null) 
            start = lastEnd.next;
            end = getKGroupEnd(start, k);
            if (end == null) 
                return head;
            
            reverse(start, end);
            lastEnd.next = end;
            lastEnd = start;
        
        return head;
    

    // 返回从start开始后的第k个节点 ...a,b,c,d,e... 比如调用getKGroupEnd(a,5) 返回e
    public static ListNode getKGroupEnd(ListNode start, int k) 
        while (--k != 0 && start != null) 
            start = start.next;
        
        return start;
    

    public static void reverse(ListNode start, ListNode end) 
        end = end.next;
        ListNode pre = null;
        ListNode cur = start;
        ListNode next = null;
        while (cur != end) 
            next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        
        start.next = end;
    

思路：

初始状态：假设k = 3

翻转第一小组链表

翻转第二小组链表

翻转第三小组链表

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