XOR链表–从头开始查找第N个节点

给定XOR链接列表和整数N ，任务是从给定XOR链接列表的末尾打印第N个节点。

例子：

Input: 4 –> 6 –> 7 –> 3, N = 1
Output: 3
Explanation: 1st node from the end is 3.

Input: 5 –> 8 –> 9, N = 4
Output: Wrong Input
Explanation: The given Xor Linked List contains only 3 nodes.

方法：请按照以下步骤解决问题：

• 使用指针curr遍历链表的前N个节点。
• 使用另一个指针(称为curr1) ，并在每次迭代后遍历链接列表，使一个节点递增curr和curr1。
• 迭代直到指针curr超过List的末尾，即NULL 。达到后，将节点curr1的值打印为所需答案。

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;
};
  
// 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 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;
    }
}
struct Node* NthNode(struct Node** head,
                     int N)
{
    int count = 0;
  
    // Stores XOR pointer
    // in current node
    struct Node* curr = *head;
    struct Node* curr1 = *head;
  
    // Stores XOR pointer of
    // in previous Node
    struct Node* prev = NULL;
    struct Node* prev1 = NULL;
  
    // Stores XOR pointer of
    // in next node
    struct Node* next;
    struct Node* next1;
  
    while (count < N && curr != NULL) {
  
        // Forward traversal
        next = XOR(prev, curr->nxp);
  
        // Update prev
        prev = curr;
  
        // Update curr
        curr = next;
  
        count++;
    }
  
    if (curr == NULL && count < N) {
        printf("Wrong Input");
        return (uintptr_t)0;
    }
    else {
        while (curr != NULL) {
  
            // Forward traversal
            next = XOR(prev,
                       curr->nxp);
            next1 = XOR(prev1,
                        curr1->nxp);
  
            // Update prev
            prev = curr;
            prev1 = curr1;
  
            // Update curr
            curr = next;
            curr1 = next1;
        }
        printf("%d", curr1->data);
    }
}
  
// 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);
  
    NthNode(&head, 3);
  
    return (0);
}

输出：

1

时间复杂度： O(N)
辅助空间： O(1)