反向单向链表中的备用 K 节点

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）。给出算法的复杂度。

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）
这种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.

C++

// C++ program to reverse alternate
// k nodes in a linked list
#include 
using namespace std;
 
/* Link list node */
class Node
{
    public:
    int data;
    Node* next;
};
 
/* Reverses alternate k nodes and
returns the pointer to the new head node */
Node *kAltReverse(Node *head, int k)
{
    Node* current = head;
    Node* next;
    Node* 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(head != NULL)
    head->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;
}
 
/* UTILITY FUNCTIONS */
/* Function to push a node */
void push(Node** head_ref, int new_data)
{
    /* allocate node */
    Node* new_node = new Node();
 
    /* put in the data */
    new_node->data = new_data;
 
    /* link the old list off the new node */
    new_node->next = (*head_ref);    
 
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
 
/* Function to print linked list */
void printList(Node *node)
{
    int count = 0;
    while(node != NULL)
    {
        cout<data<<" ";
        node = node->next;
        count++;
    }
}
 
/* Driver code*/
int main(void)
{
    /* Start with the empty list */
    Node* head = NULL;
    int i;
     
    // create a list 1->2->3->4->5...... ->20
    for(i = 20; i > 0; i--)
    push(&head, i);
 
    cout<<"Given linked list \n";
    printList(head);
    head = kAltReverse(head, 3);
 
    cout<<"\n Modified Linked list \n";
    printList(head);
    return(0);
}
 
 
// This code is contributed by rathbhupendra

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 has been contributed by Mayank Jaiswal

Python3

# Python3 program to reverse alternate
# k nodes in a linked list
import math
 
# Link list node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
# Reverses alternate k nodes and
#returns the pointer to the new head node
def kAltReverse(head, k) :
    current = head
    next = None
    prev = None
    count = 0
 
    #1) reverse first k nodes of the linked list
    while (current != None and count < k) :
        next = current.next
        current.next = prev
        prev = current
        current = next
        count = count + 1;
     
    # 2) Now head pos to the kth node.
    # So change next of head to (k+1)th node
    if(head != None):
        head.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 and current != None ):
        current = current.next
        count = count + 1
     
    # 4) Recursively call for the list
    # starting from current.next. And make
    # rest of the list as next of first node
    if(current != None):
        current.next = kAltReverse(current.next, k)
 
    # 5) prev is new head of the input list
    return prev
 
# UTILITY FUNCTIONS
# Function to push a node
def push(head_ref, new_data):
     
    # allocate node
    new_node = Node(new_data)
 
    # put in the data
    # new_node.data = new_data
 
    # link the old list off the new node
    new_node.next = head_ref
 
    # move the head to po to the new node
    head_ref = new_node
     
    return head_ref
 
# Function to print linked list
def prList(node):
    count = 0
    while(node != None):
        print(node.data, end = " ")
        node = node.next
        count = count + 1
     
# Driver code
if __name__=='__main__':
     
    # Start with the empty list
    head = None
 
    # create a list 1.2.3.4.5...... .20
    for i in range(20, 0, -1):
        head = push(head, i)
         
    print("Given linked list ")
    prList(head)
    head = kAltReverse(head, 3)
 
    print("\nModified Linked list")
    prList(head)
     
# This code is contributed by Srathore

C#

// C# program to reverse alternate
// k nodes in a linked list
using System;
class LinkedList
{
 
    static Node head;
 
    public class Node
    {
 
        public int data;
        public Node next;
 
        public 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)
        {
            Console.Write(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);
        }
        Console.WriteLine("Given Linked List :");
        list.printList(head);
        head = list.kAltReverse(head, 3);
        Console.WriteLine("");
        Console.WriteLine("Modified Linked List :");
        list.printList(head);
    }
}
 
// This code has been contributed by Arnab Kundu

Javascript

 
// JavaScript program to reverse
// alternate k nodes in a linked list
class Node
{
    constructor(d)
    {
        this.data = d;
        this.next = null;
    }
}
 
let head;
 
// Reverses alternate k nodes and returns
// the pointer to the new head node
function kAltReverse(node, k)
{
    let current = node;
    let next = null, prev = null;
    let 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;
}
 
function printList(node)
{
    while (node != null)
    {
        document.write(node.data + " ");
        node = node.next;
    }
}
 
function push(newdata)
{
    let mynode = new Node(newdata);
    mynode.next = head;
    head = mynode;
}
 
// Driver code
 
// Creating the linkedlist
for(let i = 20; i > 0; i--)
{
    push(i);
}
document.write("Given Linked List :<br>");
printList(head);
head = kAltReverse(head, 3);
 
document.write("<br>");
document.write("Modified Linked List :<br>");
printList(head);
 
// This code is contributed by rag2127
 

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 将前 k 个节点反转，然后将指针移动到前面的 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 bool 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.

C++

#include 
using namespace std;
 
/* Link list node */
class node
{
    public:
    int data;
    node* next;
};
 
/* Helper function for kAltReverse() */
node * _kAltReverse(node *node, int k, bool b);
 
/* 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 iteself */
node * _kAltReverse(node *Node, int k, bool b)
{
    if(Node == NULL)
        return NULL;
     
    int count = 1;
    node *prev = NULL;
    node *current = Node;
    node *next;
     
    /* 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 rest of the list after 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;    
    }
}
 
 
/* UTILITY FUNCTIONS */
/* Function to push a node */
void push(node** head_ref, int new_data)
{
    /* allocate node */
    node* new_node = new node();
 
    /* put in the data */
    new_node->data = new_data;
 
    /* link the old list off the new node */
    new_node->next = (*head_ref);
 
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
 
/* Function to print linked list */
void printList(node *node)
{
    int count = 0;
    while(node != NULL)
    {
        cout << node->data << " ";
        node = node->next;
        count++;
    }
}
 
// Driver Code
int main(void)
{
    /* Start with the empty list */
    node* head = NULL;
    int i;
 
    // create a list 1->2->3->4->5...... ->20
    for(i = 20; i > 0; i--)
    push(&head, i);
 
    cout << "Given linked list \n";
    printList(head);
    head = kAltReverse(head, 3);
 
    cout << "\nModified Linked list \n";
    printList(head);
    return(0);
}
 
// This code is contributed by rathbhupendra

C

#include
#include
 
/* Link list node */
struct node
{
    int data;
    struct node* next;
};
 
/* Helper function for kAltReverse() */
struct node * _kAltReverse(struct node *node, int k, bool b);
 
/* Alternatively reverses the given linked list in groups of
   given size k. */
struct node *kAltReverse(struct 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 iteself  */
struct node * _kAltReverse(struct node *node, int k, bool b)
{
   if(node == NULL)
       return NULL;
 
   int count = 1;
   struct node *prev = NULL;
   struct node  *current = node;
   struct node *next;
  
   /* 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 rest of the list after 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;      
   }
}
  
 
/* UTILITY FUNCTIONS */
/* Function to push a node */
void push(struct node** head_ref, int new_data)
{
    /* allocate node */
    struct node* new_node =
            (struct node*) malloc(sizeof(struct node));
  
    /* put in the data  */
    new_node->data  = new_data;
  
    /* link the old list off the new node */
    new_node->next = (*head_ref);
  
    /* move the head to point to the new node */
    (*head_ref)    = new_node;
}
  
/* Function to print linked list */
void printList(struct node *node)
{
    int count = 0;
    while(node != NULL)
    {
        printf("%d  ", node->data);
        node = node->next;
        count++;
    }
}
  
/* Driver program to test above function*/
int main(void)
{
    /* Start with the empty list */
    struct node* head = NULL;
    int i;
  
    // create a list 1->2->3->4->5...... ->20
    for(i = 20; i > 0; i--)
      push(&head, i);
  
    printf("\n Given linked list \n");
    printList(head);
    head = kAltReverse(head, 3);
  
    printf("\n Modified Linked list \n");
    printList(head);
  
    getchar();
    return(0);
}

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 iteself  */
    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 rest of the list after 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;
    }
 
    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 has been contributed by Mayank Jaiswal

蟒蛇3

# Python code for above algorithm
 
# Link list node
class node:
     
    def __init__(self, data):
        self.data = data
        self.next = next
         
# function to insert a node at the
# beginning of the linked list
def push(head_ref, new_data):
 
    # allocate node
    new_node = node(0)
 
    # put in the data
    new_node.data = new_data
 
    # link the old list to the new node
    new_node.next = (head_ref)
 
    # move the head to point to the new node
    (head_ref) = new_node
     
    return head_ref
     
""" Alternatively reverses the given linked list
in groups of given size k. """
def kAltReverse(head, 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 iteself """
def _kAltReverse(Node, k, b) :
 
    if(Node == None) :
        return None
     
    count = 1
    prev = None
    current = Node
    next = None
     
    """ 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 != None and count <= k) :
     
        next = current.next
     
        """ Reverse the nodes only if b is True"""
        if(b == True) :
            current.next = prev
                 
        prev = current
        current = next
        count = count + 1
     
         
    """ 3) If b is True, then node is the kth node.
        So attach rest of the list after node.
        4) After attaching, return the new head """
    if(b == True) :
     
        Node.next = _kAltReverse(current, k, not b)
        return prev        
     
    else :
        """ If b is not True, then attach
        rest of the list after prev.
        So attach rest of the list after prev """
        prev.next = _kAltReverse(current, k, not b)
        return Node    
     
""" Function to print linked list """
def printList(node) :
 
    count = 0
    while(node != None) :
     
        print( node.data, end = " ")
        node = node.next
        count = count + 1
 
# Driver Code
 
""" Start with the empty list """
head = None
i = 20
 
# create a list 1->2->3->4->5...... ->20
while(i > 0 ):
    head = push(head, i)
    i = i - 1
 
print( "Given linked list ")
printList(head)
head = kAltReverse(head, 3)
 
print( "\nModified Linked list ")
printList(head)
 
# This code is contributed by Arnab Kundu

C＃

// C# Program for converting
// singly linked list into
// circular linked list.
using System;
 
public class LinkedList
{
 
    static Node head;
 
    public class Node
    {
 
        public int data;
        public Node next;
 
        public 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)
        {
            Console.Write(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);
        }
        Console.WriteLine("Given Linked List :");
        list.printList(head);
        head = list.kAltReverse(head, 3);
        Console.WriteLine("");
        Console.WriteLine("Modified Linked List :");
        list.printList(head);
    }
}
 
// This code is contributed 29AjayKumar

Javascript

// javascript program to reverse alternate k nodes in a linked list
var head;
 
     class Node {
            constructor(val) {
                this.data = val;
                this.next = null;
            }
        }
      
 
    /*
     * Alternatively reverses the given linked list in groups of given size k.
     */
    function kAltReverse(head , 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 iteself
     */
    function _kAltReverse(node , k,  b) {
        if (node == null) {
            return null;
        }
 
        var count = 1;
var prev = null;
var current = node;
var 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 rest of the list after
         * 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;
        }
    }
 
    function printList(node) {
        while (node != null) {
            document.write(node.data + " ");
            node = node.next;
        }
    }
 
    function push(newdata) {
var mynode = new Node(newdata);
        mynode.next = head;
        head = mynode;
    }
 
     
     
 
        // Creating the linkedlist
        for (i = 20; i > 0; i--) {
            push(i);
        }
        document.write("Given Linked List :<br/>");
        printList(head);
        head = kAltReverse(head, 3);
        document.write("<br/>");
        document.write("Modified Linked List :<br/>");
        printList(head);
 
 
// This code contributed by aashish1995

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)

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