XOR 链表：反转链表的最后 K 个节点

给定一个 XOR 链表和一个正整数K ，任务是反转给定 XOR 链表中的最后K 个节点。

例子：

Input: LL: 7 <–> 6 <–> 8 <–> 11 <–> 3 <–> 1, K = 3
Output: 7<–>6<–>8<–>1<–>3<–>11

Input: LL: 7 <–> 6 <–> 8 <–> 11 <–> 3 <–> 1 <–> 2 <–> 0, K = 5
Output: 7<–>6<–>8<–>0<–>2<–>1<–>3<–>11

方法：按照以下步骤解决给定的问题：

• 反转给定的 XOR 链表。
• 现在，反转反向链表的前 K 个节点。
• 完成上述步骤后，再次反转链表，得到结果链表。

下面是上述方法的实现：

C

// C program for the above approach
  
#include 
#include 
#include 
  
// Structure of a node
// of XOR Linked List
struct Node {
  
    // Stores data value
    // of a node
    int data;
  
    // Stores XOR of previous
    // pointer and next pointer
    struct Node* nxp;
};
  
// Function to calculate
// Bitwise XOR of the two nodes
struct Node* XOR(struct Node* a,
                 struct Node* b)
{
    return (struct Node*)((uintptr_t)(a)
                          ^ (uintptr_t)(b));
}
  
// Function to insert a node with
// given value at given position
struct Node* insert(struct Node** head,
                    int value)
{
    // If XOR linked list is empty
    if (*head == NULL) {
  
        // Initialize a new Node
        struct Node* node
            = (struct Node*)malloc(
                sizeof(struct Node));
  
        // Stores data value in the node
        node->data = value;
  
        // Stores XOR of previous
        // and next pointer
        node->nxp = XOR(NULL, NULL);
  
        // Update pointer of head node
        *head = node;
    }
  
    // If the XOR linked
    // list is not empty
    else {
  
        // Stores the address
        // of the current node
        struct Node* curr = *head;
  
        // Stores the address
        // of the previous node
        struct Node* prev = NULL;
  
        // Initialize a new Node
        struct Node* node
            = (struct Node*)malloc(
                sizeof(struct Node));
  
        // Update address of current node
        curr->nxp = XOR(node,
                        XOR(
                            NULL, curr->nxp));
  
        // Update address of the new node
        node->nxp = XOR(NULL, curr);
  
        // Update the head node
        *head = node;
  
        // Update the data
        // value of current node
        node->data = value;
    }
    return *head;
}
  
// Function to print elements
// of the XOR Linked List
void printList(struct Node** head)
{
    // Stores XOR pointer
    // in the current node
    struct Node* curr = *head;
  
    // Stores XOR pointer
    // in the previous Node
    struct Node* prev = NULL;
  
    // Stores XOR pointer in the
    // next node
    struct Node* next;
  
    // Traverse XOR linked list
    while (curr != NULL) {
  
        // Print the current node
        printf("%d ", curr->data);
  
        // Forward traversal
        next = XOR(prev, curr->nxp);
  
        // Update the prev pointer
        prev = curr;
  
        // Update the curr pointer
        curr = next;
    }
}
  
// Function to reverse the linked
// list in the groups of K
struct Node* reverseK(struct Node** head,
                      int K, int len)
{
    struct Node* curr = *head;
  
    // If head is NULL
    if (curr == NULL)
        return NULL;
  
    // If the size of XOR linked
    // list is less than K
    else if (len < K)
        return *head;
    else {
  
        int count = 0;
  
        // Stores the XOR pointer
        // in the previous Node
        struct Node* prev = NULL;
  
        // Stores the XOR pointer
        // in the next node
        struct Node* next;
  
        while (count < K) {
  
            // Forward traversal
            next = XOR(prev, curr->nxp);
  
            // Update the prev pointer
            prev = curr;
  
            // Update the curr pointer
            curr = next;
  
            // Count the number of
            // nodes processed
            count++;
        }
  
        // Remove the prev node
        // from the next node
        prev->nxp = XOR(NULL,
                        XOR(prev->nxp,
                            curr));
  
        // Add the head pointer with prev
        (*head)->nxp = XOR(XOR(NULL,
                               (*head)->nxp),
                           curr);
  
        // Add the prev with the head
        if (curr != NULL)
            curr->nxp = XOR(XOR(curr->nxp,
                                prev),
                            *head);
        return prev;
    }
}
  
// Function to reverse last K nodes
// of the given XOR Linked List
void reverseLL(struct Node* head,
               int N, int K)
{
  
    // Reverse the given XOR LL
    head = reverseK(&head, N, N);
  
    // Reverse the first K nodes of
    // the XOR LL
    head = reverseK(&head, K, N);
  
    // Reverse the given XOR LL
    head = reverseK(&head, N, N);
  
    // Print the final linked list
    printList(&head);
}
  
// Driver Code
int main()
{
    // Stores number of nodes
    int N = 6;
  
    // Given XOR Linked List
  
    struct Node* head = NULL;
    insert(&head, 1);
    insert(&head, 3);
    insert(&head, 11);
    insert(&head, 8);
    insert(&head, 6);
    insert(&head, 7);
  
    int K = 3;
  
    reverseLL(head, N, K);
  
    return (0);
}

输出：

7 6 8 1 3 11

时间复杂度： O(N)
辅助空间： O(1)

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