链表问题（2）-----删除

一、题目：在单链表和双链表中删除倒数第K个节点

分别实现两个函数，一个可以删除单链表中倒数第K个节点，另一个可以删除双链表中倒数第K个节点。

要求：如果链表长度为N，时间复杂度达到O（N），额外空间复杂度达到O（1）

单链表思路：

遍历链表，每移动一步，K-1。如果链表结束，K>0 ，则不存在倒数第K个节点。如果 K == 0，则将链表的头节点删除。如果K<0，则遍历第二遍链表。

import copy
class Node:
    def __init__(self,value):
        self.value = value
        self.next = None
def delete(head,k):
    if k < 1 or not head:
        return head
    head1 = head
    while head1:      
        head1 = head1.next    
        k -= 1
    if k > 0:
        return 
    elif k == 0:
        head = head.next
    else:
        head1 = head
        k += 1
        while k != 0:
            head1 = head1.next
            k += 1
        head1.next = head1.next.next
    return head
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
k = 2
delete(head,k)
        
    

双链表的思路：

和单链表一样，只是在删除链表节点的地方考虑双链表的 last 指向。

class Node:
    def __init__(self,value):
        self.value = value
        self.last = None
        self.next = None
def delete(head,k):
    if k < 1 or not head:
        return head
    head1 = head
    while head1:      
        head1 = head1.next    
        k -= 1
    if k > 0:
        return 
    elif k == 0:
        head = head.next
        head.last = None
    else:
        head1 = head
        k += 1
        while k != 0:
            head1 = head1.next
            k += 1
        head1.next = head1.next.next
        if head1.next:
            head1.next.last = head1
    return head
head = Node(1)
head.next = Node(2)
head.last = None
head.next.next = Node(3)
head.next.last = head
head.next.next.next = Node(4)
head.next.next.last = head.next
head.next.next.next.last = head.next.next
k = 2
delete(head,k)