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class Node(object):
def __init__(self, item):
self.item = item
self.next = None
def __str__(self):
return str(self.item)
# 单向链表
class LinkedList(object):
def __init__(self):
self.head: Node = None
def is_empty(self):
return self.head is None
def length(self):
"""
O(n) 此处可以存储为一个长度变量
:return:
"""
length = 0
cur = self.head
while cur is not None:
cur = cur.next
length += 1
return length
def append(self, value: object):
"""
O(n) 将元素添加到最后 如果这里是双端链表,那么会有tail指向最后一个,能够提高到O(1)
:param value:
:return:
"""
node = Node(value)
cur = self.head
if cur is None:
self.head = node
else:
while cur.next is not None: # 只有当前节点有下一个节点, 才进行赋值
cur = cur.next
cur.next = node
def add(self, value: object):
"""
O(1) 将元素添加到头部
:param value:
:return:
"""
node = Node(value)
node.next = self.head # 将新节点的下一节点指向当前链表头部第一个节点
self.head = node # 更新当前链表的节点
def insert(self, index, value):
"""
O(1) or O(n)
:param index:
:param value:
:return:
"""
# index = index - 1
if index <= 0 or self.is_empty():
self.add(value) # 在头部插入
elif index >= self.length():
self.append(value) # 在尾部插入
else:
node = Node(value)
count = 0
cur = self.head
while cur is not None:
if index - 1 == count: # 判断是否是需要插入的位置
node.next = cur.next # 将新节点的下一节点指向当前节点(当前cur节点)的下一节点
cur.next = node # 将当前遍历节点的下一节点指向新节点
break
cur = cur.next
count += 1
def insert2(self, index, value):
"""
O(1) or O(n)
:param index:
:param value:
:return:
"""
if index <= 0 or self.is_empty():
self.add(value) # 在头部插入
elif index >= self.length():
self.append(value) # 在尾部插入
else:
node = Node(value)
count = 0
cur = self.head
while count < (index - 1):
cur = cur.next
count += 1
node.next = cur.next
cur.next = node
def reverse(self):
"""
反转链表: 顺序遍历每个节点, 将每个节点的next指针指向上一个节点
"""
cur = self.head # 当前节点指向头部
prev_node = None # 上一个节点
while cur is not None:
next_node = cur.next # 当前节点的下一个节点 必须第一步
cur.next = prev_node # 设置当前节点的下一个节点为上一节点(反转指向)
prev_node = cur # 更新变量 设置上一个节点为当前节点(下一次循环使用)
cur = next_node # 更新变量 设置当前节点为下一节点(下一次循环使用)
if cur is not None: # 最后一个节点的next必定为None, 所以这里要判断 设置head指针指向最后一个节点
self.head = cur
def remove(self, item):
cur = self.head # 双指针-当前
prev_node: Node = None # 上一个
while cur is not None:
if cur.item == item:
if prev_node is None: # 上一个节点
self.head = cur.next
else:
prev_node.next = cur.next # 删除中间元素
return True
else:
prev_node = cur
cur = cur.next
return False
def get(self, index):
cur = self.head
count = 0
while cur is not None: # 遍历节点
if count == index: # 如果次数一样
return cur.item # 返回元素
cur = cur.next
count += 1
return None
def find(self, item):
cur = self.head
count = 0
while cur is not None:
if cur.item == item:
return count
cur = cur.next
count += 1
return -1
def __iter__(self):
"""遍历"""
cur = self.head
while cur is not None:
yield cur.item
cur = cur.next
if __name__ == '__main__':
linked = LinkedList()
assert linked.is_empty() is True
linked.append(1)
linked.append(2)
linked.append(3)
linked.add(99)
# 链表长度
li = [i for i in linked]
assert li == [99, 1, 2, 3]
assert linked.length() == len(li)
# 链表反转
linked.reverse()
li = [i for i in linked]
assert li == [3, 2, 1, 99]
assert linked.length() == len(li)
# 链表插入
linked.insert(1, 100)
li = [i for i in linked]
assert li[1] == 100
# 链表删除
linked.remove(100)
li = [x for x in linked]
assert li == [3, 2, 1, 99]
# 获取元素
assert linked.get(3) == 99
assert linked.get(4) is None
# 查找位置
assert linked.find(99) == 3
assert linked.find(3) == 0
print(li)
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