和单向单端链表不同, 除了头端,尾端也会维护一个指针(能够加快在尾端添加节点的速度).
由于有着对最后一个链结点的直接引用.所以双端链表比传统链表在某些方面要方便.比如在尾部插入一个链结点.双端链表可以进行直接操作
但传统链表只能通过next节点循环找到最后链结点操作.所以双端链表适合制造队列(先进先出).
双端链表可以插入表尾，但是仍然不能方便的删除表尾，因为我们没有方法快捷地获取倒数第二个链结点，双端链表没有逆向的指针，这一点就要靠双向链表来解决了
实现参考:
class Node(object):
def __init__(self, value=None, next=None): # 这里我们 root 节点默认都是 None，所以都给了默认值
self.value = value
self.next = next
def __str__(self):
"""方便你打出来调试，复杂的代码可能需要断点调试"""
return '<Node: value: {}, next={}>'.format(self.value, self.next)
__repr__ = __str__
class LinkedList(object):
""" 链接表 ADT
[root] -> [node0] -> [node1] -> [node2]
"""
def __init__(self, maxsize=None):
"""
:param maxsize: int or None, 如果是 None，无限扩充
"""
self.maxsize = maxsize
self.root = Node() # 默认 root 节点指向 None
self.tailnode = None
self.length = 0
def __len__(self):
return self.length
def append(self, value): # O(1)
if self.maxsize is not None and len(self) >= self.maxsize:
raise Exception('LinkedList is Full')
node = Node(value) # 构造节点
tailnode = self.tailnode
if tailnode is None: # 还没有 append 过，length = 0， 追加到 root 后
self.root.next = node
else: # 否则追加到最后一个节点的后边，并更新最后一个节点是 append 的节点
tailnode.next = node
self.tailnode = 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)
if self.tailnode is None: # 如果原链表为空，插入第一个元素需要设置 tailnode
self.tailnode = node
headnode = self.root.next
self.root.next = node
node.next = headnode
self.length += 1
def __iter__(self):
for node in self.iter_node():
yield node.value
def iter_node(self):
"""遍历 从 head 节点到 tail 节点"""
curnode = self.root.next
while curnode is not self.tailnode: # 从第一个节点开始遍历
yield curnode
curnode = curnode.next # 移动到下一个节点
if curnode is not None:
yield curnode
def remove(self, value): # O(n)
""" 删除包含值的一个节点，将其前一个节点的 next 指向被查询节点的下一个即可
:param value:
"""
prevnode = self.root #
for curnode in self.iter_node():
if curnode.value == value:
prevnode.next = curnode.next
if curnode is self.tailnode: # NOTE: 注意更新 tailnode
if prevnode is self.root:
self.tailnode = None
else:
self.tailnode = prevnode
del curnode
self.length -= 1
return 1 # 表明删除成功
else:
prevnode = curnode
return -1 # 表明删除失败
def find(self, value): # O(n)
""" 查找一个节点，返回序号，从 0 开始
:param value:
"""
index = 0
for node in self.iter_node(): # 我们定义了 __iter__，这里就可以用 for 遍历它了
if node.value == value:
return index
index += 1
return -1 # 没找到
def popleft(self): # O(1)
""" 删除第一个链表节点
"""
if self.root.next is None:
raise Exception('pop from empty LinkedList')
headnode = self.root.next
self.root.next = headnode.next
self.length -= 1
value = headnode.value
if self.tailnode is headnode: # 勘误：增加单节点删除 tailnode 处理
self.tailnode = None
del headnode
return value
def clear(self):
for node in self.iter_node():
del node
self.root.next = None
self.length = 0
self.tailnode = None
def reverse(self):
"""反转链表"""
curnode = self.root.next
self.tailnode = curnode # 记得更新 tailnode，多了这个属性处理起来经常忘记
prevnode = None
while curnode:
nextnode = curnode.next
curnode.next = prevnode
if nextnode is None:
self.root.next = curnode
prevnode = curnode
curnode = nextnode
def test_linked_list():
ll = LinkedList()
ll.append(0)
ll.append(1)
ll.append(2)
ll.append(3)
print(ll)
assert len(ll) == 4
assert ll.find(2) == 2
assert ll.find(-1) == -1
assert ll.remove(0) == 1
assert ll.remove(10) == -1
assert ll.remove(2) == 1
assert len(ll) == 2
assert list(ll) == [1, 3]
assert ll.find(0) == -1
ll.appendleft(0)
assert list(ll) == [0, 1, 3]
assert len(ll) == 3
headvalue = ll.popleft()
assert headvalue == 0
assert len(ll) == 2
assert list(ll) == [1, 3]
assert ll.popleft() == 1
assert list(ll) == [3]
ll.popleft()
assert len(ll) == 0
assert ll.tailnode is None
ll.clear()
assert len(ll) == 0
assert list(ll) == []
def test_linked_list_remove():
ll = LinkedList()
ll.append(3)
ll.append(4)
ll.append(5)
ll.append(6)
ll.append(7)
ll.remove(7)
print(list(ll))
def test_single_node():
# https://github.com/PegasusWang/python_data_structures_and_algorithms/pull/21
ll = LinkedList()
ll.append(0)
ll.remove(0)
ll.appendleft(1)
assert list(ll) == [1]
def test_linked_list_reverse():
ll = LinkedList()
n = 10
for i in range(n):
ll.append(i)
ll.reverse()
assert list(ll) == list(reversed(range(n)))
def test_linked_list_append():
ll = LinkedList()
ll.appendleft(1)
ll.append(2)
assert list(ll) == [1, 2]
if __name__ == '__main__':
test_single_node()
test_linked_list()
test_linked_list_append()
test_linked_list_reverse()