C程序使用递归创建单个链接列表的副本

给定指向链接列表头节点的指针，任务是使用递归创建链接列表的副本。

范例：

Input: Head of following linked list
1->2->3->4->NULL
Output: 
Original list: 1 -> 2 -> 3 -> 4 -> NULL
Duplicate list: 1 -> 2 -> 3 -> 4 -> NULL

Input: Head of following linked list
1->2->3->4->5->NULL
Output: 
Original list: 1->2->3->4->5->NULL,
Duplicate list: 1->2->3->4->5->NULL,

方法：请按照以下步骤解决问题：

1. 基本情况： if( head == NULL )，然后返回NULL 。
2. 使用malloc()在堆中分配新节点并设置其数据。
3. 通过重复其余节点的递归设置新节点的下一个指针。
4. 返回重复节点的头指针。
5. 最后，同时打印原始链表和重复链表。

下面是上述方法的实现：

C

// C program for the above approach
#include 
#include 
  
// Node for linked list
struct Node {
    int data;
    struct Node* next;
};
  
// Function to print given linked list
void printList(struct Node* head)
{
    struct Node* ptr = head;
    while (ptr) {
  
        printf("%d -> ", ptr->data);
        ptr = ptr->next;
    }
  
    printf("NULL");
}
  
// Function to create a new node
void insert(struct Node** head_ref, int data)
{
    // Allocate the memory for new Node
    // in the heap and set its data
    struct Node* newNode
        = (struct Node*)malloc(
            sizeof(struct Node));
  
    newNode->data = data;
  
    // Set the next node pointer of the
    // new Node to point to the current
    // node of the list
    newNode->next = *head_ref;
  
    // Change the pointer of head to point
    // to the new Node
    *head_ref = newNode;
}
  
// Function to create a copy of a linked list
struct Node* copyList(struct Node* head)
{
    if (head == NULL) {
        return NULL;
    }
    else {
  
        // Allocate the memory for new Node
        // in the heap and set its data
        struct Node* newNode
            = (struct Node*)malloc(
                sizeof(struct Node));
  
        newNode->data = head->data;
  
        // Recursively set the next pointer of
        // the new Node by recurring for the
        // remaining nodes
        newNode->next = copyList(head->next);
  
        return newNode;
    }
}
  
// Function to create the new linked list
struct Node* create(int arr[], int N)
{
    // Pointer to point the head node
    // of the singly linked list
    struct Node* head_ref = NULL;
  
    // Construct the linked list
    for (int i = N - 1; i >= 0; i--) {
  
        insert(&head_ref, arr[i]);
    }
  
    // Return the head pointer
    return head_ref;
}
  
// Function to create both the lists
void printLists(struct Node* head_ref,
                struct Node* dup)
{
  
    printf("Original list: ");
  
    // Print the original linked list
    printList(head_ref);
  
    printf("\nDuplicate list: ");
  
    // Print the duplicate linked list
    printList(dup);
}
  
// Driver Code
int main(void)
{
    // Given nodes value
    int arr[] = { 1, 2, 3, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
  
    // Head of the original Linked list
    struct Node* head_ref = create(arr, N);
  
    // Head of the duplicate Linked List
    struct Node* dup = copyList(head_ref);
  
    printLists(head_ref, dup);
  
    return 0;
}

输出：

Original list: 1 -> 2 -> 3 -> 4 -> 5 -> NULL
Duplicate list: 1 -> 2 -> 3 -> 4 -> 5 -> NULL

时间复杂度： O(N)
辅助空间： O(N)

想要从精选的最佳视频中学习和练习问题，请查看《基础到高级C的C基础课程》。