用于反转单链表中的备用 K 节点的Java程序

给定一个链表，编写一个函数以有效的方式反转每个交替的 k 个节点（其中 k 是函数的输入）。给出算法的复杂性。

例子：

Inputs:   1->2->3->4->5->6->7->8->9->NULL and k = 3
Output:   3->2->1->4->5->6->9->8->7->NULL. 

方法 1（处理 2k 个节点并递归调用列表的其余部分）：
此方法基本上是本文中讨论的方法的扩展。

kAltReverse(struct node *head, int k)
  1)  Reverse first k nodes.
  2)  In the modified list head points to the kth node.  So change next 
       of head to (k+1)th node
  3)  Move the current pointer to skip next k nodes.
  4)  Call the kAltReverse() recursively for rest of the n - 2k nodes.
  5)  Return new head of the list.

Java

// Java program to reverse alternate k 
// nodes in a linked list
class LinkedList 
{
    static Node head;
  
    class Node 
    {
        int data;
        Node next;
  
        Node(int d) 
        {
            data = d;
            next = null;
        }
    }
  
    /* Reverses alternate k nodes and
       returns the pointer to the new 
       head node */
    Node kAltReverse(Node node, int k) 
    {
        Node current = node;
        Node next = null, prev = null;
        int count = 0;
  
        / *1) reverse first k nodes of the 
          linked list */
        while (current != null && count < k)  
        {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
            count++;
        }
  
        /* 2) Now head points to the kth node. 
           So change next of head to (k+1)th node*/
        if (node != null) 
        {
            node.next = current;
        }
  
        /* 3) We do not want to reverse next 
           k nodes. So move the current pointer 
           to skip next k nodes */
        count = 0;
        while (count < k - 1 && 
               current != null) 
        {
            current = current.next;
            count++;
        }
  
        /* 4) Recursively call for the list starting 
           from current->next. And make rest of the 
           list as next of first node */
        if (current != null) 
        {
            current.next = 
                    kAltReverse(current.next, k);
        }
  
        /* 5) prev is new head of the 
           input list */
        return prev;
    }
  
    void printList(Node node) 
    {
        while (node != null) 
        {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }
  
    void push(int newdata) 
    {
        Node mynode = new Node(newdata);
        mynode.next = head;
        head = mynode;
    }
  
    public static void main(String[] args) 
    {
        LinkedList list = new LinkedList();
  
        // Creating the linkedlist
        for (int i = 20; i > 0; i--) 
        {
            list.push(i);
        }
  
        System.out.println("Given Linked List :");
        list.printList(head);
        head = list.kAltReverse(head, 3);
        System.out.println("");
        System.out.println("Modified Linked List :");
        list.printList(head);
    }
}
// This code is contributed by Mayank Jaiswal

输出：

Given linked list
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Modified Linked list
3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19

时间复杂度： O(n)

方法2（处理k个节点并递归调用列表的其余部分）：
方法1反转第一个k个节点，然后将指针向前移动到k个节点。因此方法 1 使用两个 while 循环，并在一次递归调用中处理 2k 个节点。

此方法在递归调用中仅处理 k 个节点。它使用第三个布尔参数 b 来决定是反转 k 个元素还是简单地移动指针。

_kAltReverse(struct node *head, int k, bool b)
  1)  If b is true, then reverse first k nodes.
  2)  If b is false, then move the pointer k nodes ahead.
  3)  Call the kAltReverse() recursively for rest of the n - k nodes and link 
       rest of the modified list with end of first k nodes. 
  4)  Return new head of the list.

Java

// Java program to reverse alternate
// k nodes in a linked list
class LinkedList 
{
    static Node head;
  
    class Node 
    {
        int data;
        Node next;
  
        Node(int d) 
        {
            data = d;
            next = null;
        }
    }
  
    /* Alternatively reverses the given
       linked list in groups of given 
       size k. */
    Node kAltReverse(Node head, int k) 
    {
        return _kAltReverse(head, k, true);
    }
  
    /* Helper function for kAltReverse().  
       It reverses k nodes of the list only 
       if the third parameter b is passed 
       as true, otherwise moves the pointer k 
       nodes ahead and recursively calls itself  */
    Node _kAltReverse(Node node,
                      int k, boolean b) 
    {
        if (node == null) 
        {
            return null;
        }
  
        int count = 1;
        Node prev = null;
        Node current = node;
        Node next = null;
  
        /* The loop serves two purposes
           1) If b is true, then it reverses 
           the k nodes 
           2) If b is false, then it moves
           the current pointer */
        while (current != null && count <= k) 
        {
            next = current.next;
  
            /* Reverse the nodes only 
               if b is true*/
            if (b == true) 
            {
                current.next = prev;
            }
  
            prev = current;
            current = next;
            count++;
        }
  
        /* 3) If b is true, then node is the 
           kth node. So attach the rest of 
           the list after the node. 
           4) After attaching, return the new 
           head */
        if (b == true) 
        {
            node.next = 
                 _kAltReverse(current, k, !b);
            return prev;
        } 
  
        /* If b is not true, then attach rest 
           of the list after prev. So attach rest 
           of the list after prev */ 
        else 
        {
            prev.next = _kAltReverse(current, k, !b);
            return node;
        }
    }
  
    void printList(Node node) 
    {
        while (node != null) 
        {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }
  
    void push(int newdata) 
    {
        Node mynode = new Node(newdata);
        mynode.next = head;
        head = mynode;
    }
    
    // Driver code
    public static void main(String[] args) 
    {
        LinkedList list = new LinkedList();
  
        // Creating the linkedlist
        for (int i = 20; i > 0; i--) 
        {
            list.push(i);
        }
        System.out.println("Given Linked List :");
        list.printList(head);
        head = list.kAltReverse(head, 3);
        System.out.println("");
        System.out.println("Modified Linked List :");
        list.printList(head);
    }
}
// This code is contributed by Mayank Jaiswal

输出：

Given linked list
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Modified Linked list
3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19

时间复杂度： O(n)

有关详细信息，请参阅有关单链表中反向备用 K 节点的完整文章！