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：该问题可以使用给定链表的成对交换元素的概念来解决。这个想法是遍历给定的异或链表并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：

• 遍历给定的异或链表，而异或链表的当前节点和下一个节点不为空
• 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)

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