给定一个单链表和一个整数K ，任务是反转给定链表的每K个节点。
例子：
Input: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULL, K = 3
Output: 3 2 1 6 5 4 8 7
Input: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULL, K = 5
Output: 5 4 3 2 1 8 7 6
方法：本文的Set 1和Set 2中讨论了两种解决此问题的方法。在本文中，将讨论基于双端队列的方法。
- 创建双端队列。
- 将前k个节点的地址存储在双端队列中。
- 从双端队列中弹出第一个和最后一个值，然后交换这些地址处的数据值。
- 重复步骤3，直到双端队列不为空。
- 对接下来的k个节点重复步骤2，直到未到达链表的末尾。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Link list node
struct node {
int data;
struct node* next;
};
// Function to insert a node at
// the head of the linked list
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 the linked list
void printList(node* head)
{
while (head != NULL) {
cout << head->data << " ";
head = head->next;
}
}
/* Function to reverse the linked list in groups of
size k and return the pointer to the new head node. */
struct node* reverse(struct node* head, int k)
{
if (head == NULL)
return head;
// Create deque to store the address
// of the nodes of the linked list
deque q;
// Store head pointer in current to
// traverse the linked list
node* current = head;
int i;
// Iterate through the entire linked
// list by moving the current
while (current != NULL) {
i = 1;
// Store addresses of the k
// nodes in the deque
while (i <= k) {
if (current == NULL)
break;
q.push_back(current);
current = current->next;
i++;
}
/* pop first and the last value from
the deque and swap the data values at
those addresses
Do this till there exist an address in
the deque or deque is not empty*/
while (!q.empty()) {
node* front = q.front();
node* last = q.back();
swap(front->data, last->data);
// pop from the front if
// the deque is not empty
if (!q.empty())
q.pop_front();
// pop from the back if
// the deque is not empty
if (!q.empty())
q.pop_back();
}
}
return head;
}
// Driver code
int main()
{
// Start with the empty list
node* head = NULL;
// Created Linked list is
// 1->2->3->4->5->6->7->8->9->10
push(&head, 10);
push(&head, 9);
push(&head, 8);
push(&head, 7);
push(&head, 6);
push(&head, 5);
push(&head, 4);
push(&head, 3);
push(&head, 2);
push(&head, 1);
int k = 2;
// Get the new head after reversing the
// linked list in groups of size k
head = reverse(head, k);
printList(head);
return 0;
} Java
// Java implementation of the above approach
import java.util.*;
class LinkedList{
static Node head;
// Creating node class
static class Node
{
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
// Inserts a new Node at front of the list.
public void push(int new_data)
{
// Allocate the Node &
// Put in the data
Node new_node = new Node(new_data);
// Make next of new Node as head
new_node.next = head;
// Move the head to point to new Node
head = new_node;
}
// Prints content of linked list
void printList(Node node)
{
while (node != null)
{
System.out.print(node.data + " ");
node = node.next;
}
}
// Function to reverse the linked list
// in groups of size k and return the
// pointer to the new head node.
Node reverse(Node head, int k)
{
if (head == null)
return head;
// Create deque to store the address
// of the nodes of the linked list
Deque q = new ArrayDeque();
// Store head pointer in current to
// traverse the linked list
Node current = head;
int i;
// Iterate through the entire linked
// list by moving the current
while (current != null)
{
i = 1;
// Store addresses of the k
// nodes in the deque
while (i <= k)
{
if (current == null)
break;
q.addLast(current);
current = current.next;
i++;
}
// pop first and the last value from
// the deque and swap the data values at
// those addresses
// Do this till there exist an address in
// the deque or deque is not empty
while (!q.isEmpty())
{
Node front = q.peekFirst();
Node last = q.peekLast();
// Swapping front and
// last nodes
int temp = front.data;
front.data = last.data;
last.data = temp;
// pop from the front if
// the deque is not empty
if (!q.isEmpty())
q.removeFirst();
// pop from the back if
// the deque is not empty
if (!q.isEmpty())
q.removeLast();
}
}
return head;
}
// Driver code
public static void main(String[] args)
{
LinkedList list = new LinkedList();
// Created Linked list is
// 1->2->3->4->5->6->7->8->9->10
list.push(10);
list.push(9);
list.push(8);
list.push(7);
list.push(6);
list.push(5);
list.push(4);
list.push(3);
list.push(2);
list.push(1);
int k = 2;
// Get the new head after reversing the
// linked list in groups of size k
head = list.reverse(head, k);
list.printList(head);
}
}
// This code is contributed by Bindu Madhav Python3
# Python3 implementation of the approach
# Link list node
class Node:
def __init__(self):
self.data = 0
self.next = None
# Function to insert a node at
# the head of the linked list
def push(head_ref, new_data):
# Allocate node
new_node = 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
return head_ref
# Function to print the linked list
def printList( head):
while (head != None) :
print( head.data, end = " ")
head = head.next
# Function to reverse the linked list in groups of
# size k and return the pointer to the new head node.
def reverse( head, k):
if (head == None):
return head
# Create deque to store the address
# of the nodes of the linked list
q = []
# Store head pointer in current to
# traverse the linked list
current = head
i = 0
# Iterate through the entire linked
# list by moving the current
while (current != None) :
i = 1
# Store addresses of the k
# nodes in the deque
while (i <= k) :
if (current == None):
break
q.append(current)
current = current.next
i = i + 1
# pop first and the last value from
# the deque and swap the data values at
# those addresses
# Do this till there exist an address in
# the deque or deque is not empty
while (len(q) > 0):
front = q[-1]
last = q[0]
temp = front.data
front.data = last.data
last.data = temp
# pop from the front if
# the deque is not empty
if (len(q) > 0):
q.pop()
# pop from the back if
# the deque is not empty
if (len(q)):
q.pop(0)
return head
# Driver code
# Start with the empty list
head = None
# Created Linked list is
# 1.2.3.4.5.6.7.8.9.10
head = push(head, 10)
head = push(head, 9)
head = push(head, 8)
head = push(head, 7)
head = push(head, 6)
head = push(head, 5)
head = push(head, 4)
head = push(head, 3)
head = push(head, 2)
head = push(head, 1)
k = 2
# Get the new head after reversing the
# linked list in groups of size k
head = reverse(head, k)
printList(head)
# This code is contributed by Arnab Kundu输出：
2 1 4 3 6 5 8 7 10 9