Python程序删除链表的备用节点

给定一个单链表，从第二个节点开始删除它的所有备用节点。例如，如果给定的链表是 1->2->3->4->5，那么你的函数应该将它转换为 1->3->5，如果给定的链表是 1->2-> 3->4 然后将其转换为 1->3。

方法1（迭代）：
跟踪要删除的节点的前一个。首先，改变前一个节点的下一个链接，迭代移动到下一个节点。

Python3

# Python3 program to remove alternate
# nodes of a linked list
import math
 
# A linked list node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
         
# Deletes alternate nodes
# of a list starting with head
def deleteAlt(head):
    if (head == None):
        return
 
    # Initialize prev and node to
    # be deleted
    prev = head
    now = head.next
 
    while (prev != None and
           now != None):
         
        # Change next link of previous
        # node
        prev.next = now.next
 
        # Free memory
        now = None
 
        # Update prev and node
        prev = prev.next
        if (prev != None):
            now = prev.next
     
# UTILITY FUNCTIONS TO TEST
# fun1() and fun2()
# Given a reference (pointer to pointer)
# to the head of a list and an , push a
# new node on the front of the list.
def push(head_ref, new_data):
     
    # Allocate node
    new_node = Node(new_data)
 
    # Put in the data
    new_node.data = new_data
 
    # Link the old list off the
    # new node
    new_node.next = head_ref
 
    # Move the head to po to the
    # new node
    head_ref = new_node
    return head_ref
 
# Function to print nodes in a
# given linked list
def printList(node):
    while (node != None):
        print(node.data, end = " ")
        node = node.next
     
# Driver code
if __name__=='__main__':
     
    # Start with the empty list
    head = None
 
    # Using head=push() to construct
    # list 1.2.3.4.5
    head = push(head, 5)
    head = push(head, 4)
    head = push(head, 3)
    head = push(head, 2)
    head = push(head, 1)
 
    print("List before calling deleteAlt() ")
    printList(head)
 
    deleteAlt(head)
 
    print("List after calling deleteAlt() ")
    printList(head)
# This code is contributed by Srathore

Python3

# Deletes alternate nodes of a list
# starting with head
def deleteAlt(head):
    if (head == None):
        return
 
    node = head.next
 
    if (node == None):
        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 Srathore

输出：

List before calling deleteAlt() 
1 2 3 4 5 
List after calling deleteAlt() 
1 3 5

时间复杂度： O(n)，其中 n 是给定链表中的节点数。

方法2（递归）：
递归代码使用与方法 1 相同的方法。递归代码简单而简短，但会导致 O(n) 递归函数调用大小为 n 的链表。

Python3

# Deletes alternate nodes of a list
# starting with head
def deleteAlt(head):
    if (head == None):
        return
 
    node = head.next
 
    if (node == None):
        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 Srathore

时间复杂度： O(n)