重新排列链表以交替第一个和最后一个元素

给定一个链表。以交替的第一个和最后一个元素的方式排列链表。

例子：

Input : 1->2->3->4->5->6->7->8
Output :1->8->2->7->3->6->4->5

Input :10->11->15->13
Output :10->13->11->15

我们在就地重新排列给定的链表中讨论了三种不同的解决方案。
在这篇文章中，讨论了一个不同的基于 Deque 的解决方案。
方法
1）创建一个空的双端队列
2) 将链表中的所有元素插入到双端队列中
3）以交替方式将元素从双端队列插入回链表，即先然后后，依此类推

C++

// CPP program to rearrange a linked list in given manner
#include 
using namespace std;
 
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
 
/* Function to reverse the linked list */
void arrange(struct Node* head)
{
    struct Node* temp = head;
    deque d; // defining a deque
 
    // push all the elements of linked list in to deque
    while (temp != NULL) {
        d.push_back(temp->data);
        temp = temp->next;
    }
 
    // Alternatively push the first and last elements
    // from deque to back to the linked list and pop
    int i = 0;
    temp = head;
    while (!d.empty()) {
        if (i % 2 == 0) {
            temp->data = d.front();
            d.pop_front();
        }
        else {
            temp->data = d.back();
            d.pop_back();
        }
        i++;
        temp = temp->next; // increase temp
    }
}
 
/*UTILITY FUNCTIONS*/
/* Push a node to linked list. Note that this function
changes the head */
void push(struct Node** head_ref, char 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 pochar to the new node */
    (*head_ref) = new_node;
}
 
// printing the linked list
void printList(struct Node* head)
{
    struct Node* temp = head;
    while (temp != NULL) {
        printf("%d  ", temp->data);
        temp = temp->next;
    }
}
 
/* Drier program to test above function*/
int main()
{
    // Let us create linked list 1->2->3->4
    struct Node* head = NULL;
 
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
    cout << "Given linked list\t";
    printList(head);
    arrange(head);
    cout << "\nAfter rearrangement\t";
    printList(head);
    return 0;
}

Java

// Java program to rearrange a linked list in given manner
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
    /* Link list node */
    static class Node
    {
        int data;
        Node next;
        Node(int data)
        {
            this.data = data;
            next = null;
        }
    }
     
    // printing the linked list
    static void printList(Node head)
    {
        Node temp = head;
        while (temp != null)
        {
            System.out.print(temp.data + " ");
            temp = temp.next;
        }
    }
 
    /* Function to reverse the linked list */
    static void arrange(Node head)
    {
        // defining a deque
        Deque deque = new ArrayDeque<>();
        Node temp = head;
         
        // push all the elements of linked list in to deque
        while(temp != null)
        {
            deque.addLast(temp.data);
            temp = temp.next;
        }
        temp = head;
        int i = 0;
         
        // Alternatively push the first and last elements
        // from deque to back to the linked list and pop
        while(!deque.isEmpty())
        {
            if(i % 2 == 0)
            {
                temp.data = deque.removeFirst();
            }
            else
            {
                temp.data = deque.removeLast();
            }
            i++;
            temp = temp.next;
        }
    }
 
    // Driver code
    public static void main (String[] args)
    {
        // Let us create linked list 1->2->3->4->5
        Node head = null;
         
        head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);
        head.next.next.next.next = new Node(5);
 
        System.out.println("Given linked list");
        printList(head);
        arrange(head);
        System.out.println("\nAfter rearrangement");
        printList(head);
    }
}
 
// This code is contributed by nobody_cares.

Python

# Python program to rearrange
# a linked list in given manner
 
# Link list node
class Node:
     
    def __init__(self, data):
        self.data = data
        self.next = None
 
# Function to reverse the linked list
def arrange( head):
 
    temp = head
     
    # defining a deque
    d = []
     
    # push all the elements of linked list in to deque
    while (temp != None) :
        d.append(temp.data)
        temp = temp.next
     
    # Alternatively push the first and last elements
    # from deque to back to the linked list and pop
    i = 0
    temp = head
    while (len(d) > 0) :
        if (i % 2 == 0) :
            temp.data = d[0]
            d.pop(0)
         
        else :
            temp.data = d[-1]
            d.pop()
         
        i = i + 1
 
        # increase temp
        temp = temp.next
         
    return head
     
# UTILITY FUNCTIONS
# Push a node to linked list. Note that this function
# changes the head
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 off the new node
    new_node.next = (head_ref)
 
    # move the head to pochar to the new node
    (head_ref) = new_node
    return head_ref
 
# printing the linked list
def printList( head):
 
    temp = head
    while (temp != None) :
        print( temp.data,end=" ")
        temp = temp.next
 
# Driver program to test above function
 
# Let us create linked list 1.2.3.4
head = None
 
head = push(head, 5)
head = push(head, 4)
head = push(head, 3)
head = push(head, 2)
head = push(head, 1)
print("Given linked list\t")
printList(head)
head = arrange(head)
print( "\nAfter rearrangement\t")
printList(head)
 
# This code is contributed by Arnab Kundu

Javascript

 
// Javascript program to rearrange a linked list in given manner
 
/* Link list node */
class Node {
 
    constructor()
    {
        this.data = 0;
        this.next = null;
    }
};
 
/* Function to reverse the linked list */
function arrange( head)
{
    var temp = head;
    var d = []; // defining a deque
 
    // push all the elements of linked list in to deque
    while (temp != null) {
        d.push(temp.data);
        temp = temp.next;
    }
 
    // Alternatively push the first and last elements
    // from deque to back to the linked list and pop
    var i = 0;
    temp = head;
    while (d.length!=0) {
        if (i % 2 == 0) {
            temp.data = d[0];
            d.shift();
        }
        else {
            temp.data = d[d.length-1];
            d.pop();
        }
        i++;
        temp = temp.next; // increase temp
    }
}
 
/*UTILITY FUNCTIONS*/
/* Push a node to linked list. Note that this function
changes the head */
function push(head_ref, new_data)
{
    /* allocate node */
    var 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 pochar to the new node */
    (head_ref) = new_node;
 
    return head_ref;
}
 
// printing the linked list
function printList(head)
{
    var temp = head;
    while (temp != null) {
        document.write(temp.data+ " ");
        temp = temp.next;
    }
}
 
/* Drier program to test above function*/
// Let us create linked list 1.2.3.4
var head = null;
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);
document.write( "Given linked list  ");
printList(head);
arrange(head);
document.write( "<br>After rearrangement  ");
printList(head);
 
 

输出：

Given linked list    1 2 3 4 5 
After rearrangement    1 5 2 4 3

时间复杂度： O(n)

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