C++程序 删除链表的交替节点

给定一个单链表，从第二个节点开始删除其所有交替节点。例如，如果给定的链表是1->2->3->4->5，则您的函数应将其转换为1->3->5，如果给定的链表是1->2->3->4，则将其转换为1->3。

Method 1 (Iterative):

要删除的节点的前一个节点。首先，更改前一个节点的下一个链接，并迭代移动到下一个节点。

// C++ program to remove alternate
// nodes of a linked list
#include <bits/stdc++.h>
using namespace std;
 
// A linked list node
class Node
{
    public:
    int data;
    Node *next;
};
 
/* Deletes alternate nodes
   of a list starting with head */
void deleteAlt(Node *head)
{
    if (head == NULL)
        return;
 
    /* Initialize prev and node
       to be deleted */
    Node *prev = head;
    Node *node = head->next;
 
    while (prev != NULL &&
           node != NULL)
    {
        /* Change next link of previous
           node */
        prev->next = node->next;
 
        // Update prev and node
        prev = prev->next;
        if (prev != NULL)
            node = prev->next;
    }
}
 
/* UTILITY FUNCTIONS TO TEST
   fun1() and fun2() */
/* Given a reference (pointer to pointer)
   to the head of a list and an int, push
   a new node on the front of the list. */
void push(Node** head_ref,
          int new_data)
{
    /* Allocate node */
    Node* new_node = new Node();
 
    /* Put in the data */
    new_node->data = new_data;
 
    /* Link the old list of the
       new node */
    new_node->next = (*head_ref);
 
    /* Move the head to point to the
       new node */
    (*head_ref) = new_node;
}
 
/* Function to print nodes in a
   given linked list */
void printList(Node *node)
{
    while (node != NULL)
    {
        cout << node->data << " ";
        node = node->next;
    }
}
 
// Driver code
int main()
{
    // Start with the empty list
    Node* head = NULL;
 
    /* Using push() to construct list
       1->2->3->4->5 */
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
 
    cout << "List before calling deleteAlt() ";
    printList(head);
 
    deleteAlt(head);
 
    cout << "List after calling deleteAlt() ";
    printList(head);
 
    return 0;
}
// This code is contributed by rathbhupendra```  

输出：

在调用deleteAlt()之前的列表为
1 2 3 4 5 
调用deleteAlt()之后的列表为
1 3 5 

时间复杂度： O(n)，其中n是给定链表中的节点数。

辅助空间 ：O(1)，因为它使用了常数空间

方法2（递归）：

递归代码使用方法1相同的方法。递归代码简单而短，但对于大小为n的链表，会导致O(n)的递归函数调用。

/* Deletes alternate nodes of a list
   starting with head */
void deleteAlt(Node *head)
{
    if (head == NULL)
        return;
 
    Node *node = head->next;
 
    if (node == NULL)
        return;
 
    // Change the next link of head
    head->next = node->next;
 
    // Free memory allocated for node
    free(node);
 
    /* Recursively call for the new
       next of head */
    deleteAlt(head->next);
}
// This code is contributed by rathbhupendra```  

时间复杂度： O(n)

辅助空间 ：因使用递归，对于函数调用堆栈的空间复杂度为O(n)