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:
- 基本情况: if( head == NULL ),然后返回NULL 。
- 使用malloc()在堆中分配新节点并设置其数据。
- 通过重复其余节点的递归设置新节点的下一个指针。
- 返回重复节点的头指针。
- 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《基础知识到高级C的C基础课程》。