用于旋转链表的 C++ 程序

给定一个单链表，将链表逆时针旋转 k 个节点。其中 k 是给定的正整数。例如，如果给定的链表是 10->20->30->40->50->60，k 为 4，则链表应修改为 50->60->10->20->30- >40。假设 k 小于链表中的节点数。

方法一：
要旋转链表，我们需要将第 k 个节点的 next 更改为 NULL，将最后一个节点的 next 更改为前一个 head 节点，最后将 head 更改为第 (k+1) 个节点。所以我们需要掌握三个节点：第k个节点、第(k+1)个节点和最后一个节点。
从头开始遍历列表并在第 k 个节点处停止。存储指向第 k 个节点的指针。我们可以使用 kthNode->next 获得第 (k+1) 个节点。继续遍历直到结束并存储指向最后一个节点的指针。最后，如上所述更改指针。

下图显示了如何在代码中使用旋转函数：

C++

// C++ program to rotate a 
// linked list counter clock wise
#include 
using namespace std;
  
// Link list node 
class Node 
{
    public:
    int data;
    Node* next;
};
  
// This function rotates a linked list
// counter-clockwise and updates the
// head. The function assumes that k is
// smaller than size of linked list.
// It doesn't modify the list if
// k is greater than or equal to size
void rotate(Node** head_ref, int k)
{
    if (k == 0)
        return;
  
    // Let us understand the below
    // code for example k = 4 and
    // list = 10->20->30->40->50->60.
    Node* current = *head_ref;
  
    // current will either point to
    // kth or NULL after this loop.
    // current will point to node
    // 40 in the above example
    int count = 1;
    while (count < k && 
           current != NULL) 
    {
        current = current->next;
        count++;
    }
  
    // If current is NULL, k is greater 
    // than or equal to count of nodes 
    // in linked list. Don't change the 
    // list in this case
    if (current == NULL)
        return;
  
    // current points to kth node.
    // Store it in a variable. kthNode
    // points to node 40 in the above 
    // example
    Node* kthNode = current;
  
    // current will point to
    // last node after this loop
    // current will point to
    // node 60 in the above example
    while (current->next != NULL)
        current = current->next;
  
    // Change next of last node to 
    // previous head. Next of 60 is 
    // now changed to node 10
    current->next = *head_ref;
  
    // Change head to (k+1)th node
    // head is now changed to node 50
    *head_ref = kthNode->next;
  
    // Change next of kth node to NULL
    // next of 40 is now NULL
    kthNode->next = NULL;
}
  
// 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)
{
    while (node != NULL) 
    {
        cout << node->data << " ";
        node = node->next;
    }
}
  
// Driver code
int main(void)
{
    // Start with the empty list 
    Node* head = NULL;
  
    // Create a list 
    // 10->20->30->40->50->60
    for (int i = 60; i > 0; i -= 10)
        push(&head, i);
  
    cout << "Given linked list ";
    printList(head);
    rotate(&head, 4);
  
    cout << "Rotated Linked list ";
    printList(head);
  
    return (0);
}
// This code is contributed by rathbhupendra

输出：

Given linked list
10  20  30  40  50  60
Rotated Linked list
50  60  10  20  30  40

方法二：
要将链表旋转k，我们可以先使链表循环，然后从头节点向前移动k-1步，使第(k-1)个节点的next为空，并以第k个节点为头。

C++

// C++ program to rotate a 
// linked list counter clock wise
#include 
using namespace std;
  
// Link list node 
class Node 
{
    public:
    int data;
    Node* next;
};
  
// This function rotates a linked list
// counter-clockwise and updates the
// head. The function assumes that k is
// smaller than size of linked list.
void rotate(Node** head_ref, int k)
{
    if (k == 0)
        return;
  
    // Let us understand the below
    // code for example k = 4 and
    // list = 10->20->30->40->50->60.
    Node* current = *head_ref;
  
    // Traverse till the end.
    while (current->next != NULL)
        current = current->next;
  
    current->next = *head_ref;
    current = *head_ref;
  
    // Traverse the linked list to 
    // k-1 position which will be 
    // last element for rotated array.
    for (int i = 0; i < k - 1; i++)
        current = current->next;
  
    // Update the head_ref and last 
    // element pointer to NULL
    *head_ref = current->next;
    current->next = NULL;
}
  
// 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)
{
    while (node != NULL) 
    {
        cout << node->data << " ";
        node = node->next;
    }
}
  
// Driver code
int main(void)
{
    // Start with the empty list 
    Node* head = NULL;
  
    // Create a list 
    // 10->20->30->40->50->60
    for (int i = 60; i > 0; i -= 10)
        push(&head, i);
  
    cout << "Given linked list ";
    printList(head);
    rotate(&head, 4);
  
    cout << "Rotated Linked list ";
    printList(head);
  
    return (0);
}
// This code is contributed by pkurada

输出：

Given linked list 
10 20 30 40 50 60 
Rotated Linked list 
50 60 10 20 30 40

有关详细信息，请参阅有关旋转链接列表的完整文章！