XOR链表–给定链表的成对交换元素

给定一个XOR链表，任务是成对交换给定XOR链表的元素。

例子：

Input: 4 7 9 7
Output: 7 4 7 9
Explanation:
First pair of nodes are swapped to formed the sequence {4, 7} and second pair of nodes are swapped to form the sequence {9, 7}.

Input: 1 2 3
Output: 2 1 3

方法：可以使用给定链表的成对交换元素的概念来解决该问题。这个想法是遍历给定的异或链表，选择两个相邻的节点对并交换它们。请按照以下步骤解决问题：

• 遍历给定的xor链表，而xor链表的当前节点和下一个节点不为null
• 在每次迭代中，将当前节点的数据与下一个节点交换，并更新下一个节点和当前节点的指针。
• 最后，打印异或链接的列表

下面是上述方法的实现：

C

// C program to implement
// the above approach
#include 
#include 
#include 
  
// Structure of a node
// in XOR linked list
struct Node {
  
    // Stores data value
    // of a node
    int data;
  
    // Stores XOR of previous
    // pointer and next pointer
    struct Node* nxp;
};
  
void pairwiseswap(struct Node**);
void swap(int*, int*);
  
// Function to find the XOR of 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 current node
        struct Node* curr = *head;
  
        // Stores the address
        // of previous node
        struct Node* prev = NULL;
  
        // Initialize a new Node
        struct Node* node
            = (struct Node*)malloc(
                sizeof(struct Node));
  
        // Update curr node address
        curr->nxp = XOR(node,
                        XOR(NULL, curr->nxp));
  
        // Update new node address
        node->nxp = XOR(NULL, curr);
  
        // Update head
        *head = node;
  
        // Update data value of
        // current node
        node->data = value;
    }
    return *head;
}
  
// Function to pairwise swap the nodes
// of the XOR-linked list
void pairwiseswap(struct Node** head)
{
  
    // Stores pointer of current node
    struct Node* curr = (*head);
  
    // Stores pointer of previous node
    struct Node* prev = NULL;
  
    // Stores pointer of next node
    struct Node* next = NULL;
  
    // Traverse the XOR-linked list
    while (curr != NULL
           && curr->nxp != XOR(prev, NULL)) {
  
        next = XOR(prev, curr->nxp);
        prev = curr;
        curr = next;
  
        // Swap the elements
        swap(&prev->data, &curr->data);
  
        // Update next
        next = XOR(prev, curr->nxp);
  
        // Update prev
        prev = curr;
  
        // Update curr
        curr = next;
    }
}
  
// Function to swap two numbers
void swap(int* a, int* b)
{
    int temp;
    temp = *a;
    *a = *b;
    *b = temp;
}
  
// Function to print elements of
// the XOR Linked List
void printList(struct Node** head)
{
    // Stores XOR pointer
    // in current node
    struct Node* curr = *head;
  
    // Stores XOR pointer of
    // in previous Node
    struct Node* prev = NULL;
  
    // Stores XOR pointer of
    // in next node
    struct Node* next;
  
    // Traverse XOR linked list
    while (curr != NULL) {
  
        // Print current node
        printf("%d ", curr->data);
  
        // Forward traversal
        next = XOR(prev, curr->nxp);
  
        // Update prev
        prev = curr;
  
        // Update curr
        curr = next;
    }
}
  
// Driver Code
int main()
{
  
    /* Create following XOR Linked List
    head -->7 –> 6 –>8 –> 11 –> 3 –> 1 –> 2 –> 0*/
    struct Node* head = NULL;
  
    insert(&head, 0);
    insert(&head, 2);
    insert(&head, 1);
    insert(&head, 3);
    insert(&head, 11);
    insert(&head, 8);
    insert(&head, 6);
    insert(&head, 7);
  
    // Swap the elements pairwise
    pairwiseswap(&head);
  
    // Print the new list
    printList(&head);
  
    return (0);
}

输出：

6 7 11 8 1 3 0 2

时间复杂度： O(N)
辅助空间： O(N)