Python程序删除链表的M个节点后的N个节点

给定一个链表和两个整数 M 和 N。遍历链表，保留 M 个节点，然后删除下一个 N 个节点，继续相同直到链表结束。
难度级别：新手
例子：

Input:
M = 2, N = 2
Linked List: 1->2->3->4->5->6->7->8
Output:
Linked List: 1->2->5->6

Input:
M = 3, N = 2
Linked List: 1->2->3->4->5->6->7->8->9->10
Output:
Linked List: 1->2->3->6->7->8

Input:
M = 1, N = 1
Linked List: 1->2->3->4->5->6->7->8->9->10
Output:
Linked List: 1->3->5->7->9

问题的主要部分是保持节点之间的正确链接，确保处理所有极端情况。以下是函数skipMdeleteN() 的 C 实现，它跳过 M 个节点并删除 N 个节点直到列表末尾。假设 M 不能为 0。

Python

# Python program to delete M nodes
# after N nodes
# Node class
class Node:
 
    # Constructor to initialize
    # the node object
    def __init__(self, data):
        self.data = data
        self.next = None
 
class LinkedList:
 
    # Function to initialize head
    def __init__(self):
        self.head = None
 
    # Function to insert a new node
    # at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node
 
    # Utility function to print the
    # linked LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print temp.data,
            temp = temp.next
 
    def skipMdeleteN(self, M, N):
        curr = self.head
         
        # The main loop that traverses
        # through the whole list
        while(curr):
            # Skip M nodes
            for count in range(1, M):
                if curr is None:
                    return
                curr = curr.next
                     
            if curr is None :
                return
 
            # Start from next node and delete
            # N nodes
            t = curr.next
            for count in range(1, N+1):
                if t is None:
                    break
                t = t.next
     
            # Link the previous list with
            # remaining nodes
            curr.next = t
 
            # Set Current pointer for next
            # iteration
            curr = t
 
# Driver code
 
# Create following linked list
# 1->2->3->4->5->6->7->8->9->10
llist = LinkedList()
M = 2
N = 3
llist.push(10)
llist.push(9)
llist.push(8)
llist.push(7)
llist.push(6)
llist.push(5)
llist.push(4)
llist.push(3)
llist.push(2)
llist.push(1)
 
print "M = %d, N = %d
Given Linked List is:" %(M, N)
llist.printList()
print
 
llist.skipMdeleteN(M, N)
 
print "Linked list after deletion is"
llist.printList()
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

输出：

M = 2, N = 3
Given Linked list is :
1 2 3 4 5 6 7 8 9 10
Linked list after deletion is :
1 2 6 7

时间复杂度：
O(n) 其中 n 是链表中的节点数。

辅助空间：O(1)

详情请参考完整文章删除链表的M个节点后的N个节点！