#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @File : 拓展单向链表.py
# @Date : 2019/3/26 0026
# @Contact : [email protected]
# @Author: DeepMan
"""
这一节主要介绍链表的其他操作 包括交叉 环形判断。等
"""
class Node:
def __init__(self, initdata=None):
self.data = initdata
self.next = None
head = Node()
# 1 2 3 4 5 6 7 8 9 顺序输入
# 9->8->7->6->5->4->3->2->1->None 这就是头插法
for i in range(1, 10):
s = Node(i)
temp = head.next
head.next = s
s.next = temp
# cur = head.next
# while cur.next != None:
# print(cur.data,end='->')
# cur = cur.next
"""
链表逆序**********************************************************************
"""
# 那怎么实现上述链表
# head->9->8->7->6->5->4->3->2->1->None 的逆序呢
# 变为 None<-9<-8<-7<-6<-5<-4<-3<-2<-1
# print()
# print("后面是进行逆序")
# NewList = Node()
#
# cur = head.next
# 同样可以使用头插入的方法
# while cur is not None:
# newnode = Node(cur.data) # 创建新节点用于接收读取到的数据
#
# newnode.next = NewList.next
# NewList.next = newnode
#
# cur = cur.next # 移动指针将所有数据遍历出来
# cur = NewList.next
# while cur != None:
# print(cur.data, end="->")
# cur = cur.next
# 最终结果：
# 9->8->7->6->5->4->3->2->
# 后面是进行逆序
# 1->2->3->4->5->6->7->8->9->
class ListNode(object):
def __init__(self, val=None):
self.val = val
self.next = None
def build_l(nums):
head = ListNode(nums[0]) # 头节点
cur = head # 移动节点
for i in nums[1:]:
cur.next = ListNode(i)
cur = cur.next # 指向 到 下一个节点
return head # 只要有了head 就可以将链表找到
def build_recurrent_list(nums): # 单向循环链表
head = ListNode(nums[0]) # 头节点
cur = head # 移动节点
mid_loop = head # 在由结尾到中间之间成环 在i= 5 的时候成Loop
for i in nums[1:]:
cur.next = ListNode(i)
cur = cur.next # 指向 到 下一个节点
if i == 12:
mid_loop = cur # 取 中间一个节点
# cur.next = head # 这是形成完整的环
# 还有一种是中间有一个环 （两者取其一）
cur.next = mid_loop
#
return head # 只要有了head 就可以将链表找到
def build_interval_list(nums): # 交叉链表
head = ListNode(nums[0]) # 头节点
cur = head
mid_interval = head
for i in nums[1:]:
cur.next = ListNode(i)
cur = cur.next
if i == 18:
mid_interval = cur
second_list_head = ListNode(val=19)
curr = second_list_head
for i in range(20, 50):
curr.next = ListNode(i)
curr = curr.next
curr.next = mid_interval
return head, second_list_head
"""
快慢指针实现循环链表的判断****************************************************
"""
# num = [1,2,3,4,5,6]
# 首先我们创建一个链表 测试的时候我们用pytest
# linkList = build_l(num)
# 测试是否是循环链表
def is_Loop_List_or_Not(linkList):
"""
测试是否是循环链表 打印输出值
:param linkList:
:return:
"""
cur = linkList
while cur is not None:
print(cur.val)
cur = cur.next
def isExitsLoop(linkList):
"""
利用快慢指针进行判断 当两者相遇的时候则有环
:param linkList:
:return:
"""
fast = linkList
slow = linkList
while(fast and fast.next):
slow = slow.next
fast = fast.next.next
if slow == fast:
print("break")
return True
return False
# print(isExitsLoop(linkList))
"""
快慢指针实现中位数查找****************************************************
"""
def middleNode1(linkList):
fast = linkList
slow = linkList
while fast.next is not None and fast.next.next is not None:
fast = fast.next.next
print(fast.val)
slow = slow.next
if fast.next is not None:
return slow.next.val # 偶数的时候返回 下节点
else:
return slow.val
# *************************
def middleNode(linkList):
fast = linkList
slow = linkList
while fast and fast.next:
fast = fast.next.next
slow = slow.next
return slow.val
num = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
# 首先我们创建一个链表 测试的时候我们用pytest
# linkList = build_recurrent_list(num)
# is_Loop_List_or_Not(linkList) # 观察是否成环
def findEntryNode(linklist): # 成环检测
fast, slow = linklist, linklist
while True:
fast = fast.next.next
slow = slow.next
if slow == fast: # 表示相遇
# 将 fast 移动到head 然后以步长为2 进行移动 此时可以将再次相遇的节点返回
newfast = linklist
break
while True:
newfast = newfast.next
slow = slow.next
if slow == newfast:
return slow.val
# print(findEntryNode(linkList))
head, second_list = build_interval_list(num)
def find_interval_in_doubleList(list1, list2):
newlist1, newlist2 = list1, list2
#将第一条链成环
cur1 = newlist1
count1 = 0
while cur1.next != None:
count1 +=1
cur1 = cur1.next
cur1.next = newlist1
# 进一步检查 是否 有入口 并返回入口序号
jiaodian = findEntryNode(newlist2)
print(jiaodian)
# find_interval_in_doubleList(head, second_list)
# 单链表的逆转 k个
def singleListinverse(k, head):
myhead = ListNode()
myhead.next = head
# 首先 定义三个指针
new = myhead.next
old = myhead.next.next
count = 1
while count < k:
temp = old.next # 防止链表的丢失
old.next = new
# 然后进行移位操作
new = old
old = temp
count += 1
myhead.next.next = old
return new
linklist = build_l(num)
pp = singleListinverse(1, linklist)
is_Loop_List_or_Not(pp)
