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stack.py
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# -*- coding: utf-8 -*-
# NOTE: 这里拷贝的 double_link_list.py 里的代码
from collections import deque
class Node(object):
def __init__(self, value=None, prev=None, next=None):
self.value, self.prev, self.next = value, prev, next
class CircularDoubleLinkedList(object):
"""循环双端链表 ADT
多了个循环其实就是把 root 的 prev 指向 tail 节点,串起来
"""
def __init__(self, maxsize=None):
self.maxsize = maxsize
node = Node()
node.next, node.prev = node, node
self.root = node
self.length = 0
def __len__(self):
return self.length
def headnode(self):
return self.root.next
def tailnode(self):
return self.root.prev
def append(self, value): # O(1), 你发现一般不用 for 循环的就是 O(1),有限个步骤
if self.maxsize is not None and len(self) >= self.maxsize:
raise Exception('LinkedList is Full')
node = Node(value=value)
tailnode = self.tailnode() or self.root
tailnode.next = node
node.prev = tailnode
node.next = self.root
self.root.prev = node
self.length += 1
def appendleft(self, value):
if self.maxsize is not None and len(self) >= self.maxsize:
raise Exception('LinkedList is Full')
node = Node(value=value)
if self.root.next is self.root: # empty
node.next = self.root
node.prev = self.root
self.root.next = node
self.root.prev = node
else:
node.prev = self.root
headnode = self.root.next
node.next = headnode
headnode.prev = node
self.root.next = node
self.length += 1
def remove(self, node): # O(1),传入node 而不是 value 我们就能实现 O(1) 删除
"""remove
:param node # 在 lru_cache 里实际上根据key 保存了整个node:
"""
if node is self.root:
return
else: #
node.prev.next = node.next
node.next.prev = node.prev
self.length -= 1
return node
def iter_node(self):
if self.root.next is self.root:
return
curnode = self.root.next
while curnode.next is not self.root:
yield curnode
curnode = curnode.next
yield curnode
def __iter__(self):
for node in self.iter_node():
yield node.value
def iter_node_reverse(self):
"""相比单链表独有的反序遍历"""
if self.root.prev is self.root:
return
curnode = self.root.prev
while curnode.prev is not self.root:
yield curnode
curnode = curnode.prev
yield curnode
############################################################
# 分割线,下边是本章 内容实现
############################################################
class Deque(CircularDoubleLinkedList): # 注意这里我们用到了继承,嗯,貌似我说过不会用啥 OOP 特性的,抱歉
def pop(self):
"""删除尾节点"""
if len(self) == 0:
raise Exception('empty')
tailnode = self.tailnode()
value = tailnode.value
self.remove(tailnode)
return value
def popleft(self):
if len(self) == 0:
raise Exception('empty')
headnode = self.headnode()
value = headnode.value
self.remove(headnode)
return value
def test_deque():
dq = Deque()
dq.append(1)
dq.append(2)
assert list(dq) == [1, 2]
dq.appendleft(0)
assert list(dq) == [0, 1, 2]
dq.pop()
assert list(dq) == [0, 1]
dq.popleft()
assert list(dq) == [1]
dq.pop()
assert len(dq) == 0
class Stack(object):
def __init__(self):
self.deque = Deque() # 你可以很容易替换为 python 内置的 collections.deque
def push(self, value):
self.deque.append(value)
def pop(self):
return self.deque.pop()
class Stack2(object):
def __init__(self):
self._deque = deque()
def push(self, value):
return self._deque.append(value)
def pop(self):
return self._deque.pop()
def empty(self):
return len(self._deque) == 0
def test_stack():
s = Stack()
s.push(0)
s.push(1)
s.push(2)
assert s.pop() == 2
assert s.pop() == 1
assert s.pop() == 0
import pytest # pip install pytest
with pytest.raises(Exception) as excinfo: # 我们来测试是否真的抛出了异常
s.pop()
assert 'empty' in str(excinfo.value)
if __name__ == '__main__':
test_stack()