if b not in self.servers: self.servers[b] = Node(b) self.servers[b].next.append(self.servers[a]) # a依赖b且b是新增节点,添加到头节点列表 self.heads.append(self.servers[b])
时间: 2024-03-04 10:53:58 浏览: 19
这是 Server 类中的 add 方法的后半部分,用于处理节点之间的依赖关系。如果 b 不在服务器列表中,那么就创建一个新的节点,并将其添加到服务器列表中。然后,将 b 添加到 a 的 next 属性中,表示 a 依赖于 b。
同时,如果 b 是新添加的节点,那么就将其添加到头节点列表中,因为它目前没有任何依赖关系,是一个合法的头节点。这样,当后续添加新的节点时,就可以根据头节点列表中的节点来构建服务器链表。
相关问题
约瑟夫环改错class Node: def __init__(self,data): self.data=data self.next=Noneclass linklist: def __init__(self): self.head=None self.data=None def isEmpty(self): if self.head: return False else: return True def length(self): if self.isEmpty(): return 0 else: t = self.head n = 1 while t.next: if t.next == self.head: break t = t.next n = n + 1 return n def addhead(self,data): node = Node(data) if self.isEmpty(): self.head = node self.tail = self.head else: node.next = self.head self.head = node self.tail.next = self.head def addtail(self,data): node=Node(data) if self.isEmpty(): self.addhead(data) else: t=self.head n=1 l=self.length() while n<l: n=n+1 t=t.next t.next=node node.next=self.head self.tail=node def delete(self,index): if self.isEmpty(): print("The linked list is empty") else: t = self.head l = self.length() if index == 0: self.head = t.next self.tail.next = self.head elif index == l - 1: n = 1 while n < l - 1: t = t.next n = n + 1 t.next = self.head self.tail = t elif index > l - 1: print("Out of range") elif index < 0: print("Wrong operation") else: n = 1 while n < index - 1: t = t.next n = n + 1 a = t.next.next t.next = a def insert(self,data,index): l = self.length() if index == 0 or self.isEmpty(): self.addhead(data) elif index >= l: self.addtail(data) else: node = Node(data) t = self.head n = 1 while n < index - 1: t = t.next n = n + 1 a = t.next t.next = node node.next = a def search(self,a): t=self.head for i in range(a): t=t.next return t.data def form(self,datalist): self.addhead(datalist[0]) for i in range(1,len(datalist)): self.addtail(datalist[i]) t = self.head while t.next != self.head: t = t.nextn,p=map(int,input().split(' '))data=[]p=p-1for i in range(1,n+1): data.append(i)print(data)datalist=[]for i in range(len(data)): datalist.append(int(data[i]))link=linklist()link.form(datalist)a=pb=[]while link.length()>0: b.append(link.search(a)) link.delete(a) a=a+p while a>=link.length(): a=a-link.length()print(b)
改错如下:
```python
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkList:
def __init__(self):
self.head = None
self.tail = None
def isEmpty(self):
if self.head:
return False
else:
return True
def length(self):
if self.isEmpty():
return 0
else:
t = self.head
n = 1
while t.next:
if t.next == self.head:
break
t = t.next
n += 1
return n
def addhead(self, data):
node = Node(data)
if self.isEmpty():
self.head = node
self.tail = self.head
else:
node.next = self.head
self.head = node
self.tail.next = self.head
def addtail(self, data):
node = Node(data)
if self.isEmpty():
self.addhead(data)
else:
t = self.head
n = 1
l = self.length()
while n < l:
n += 1
t = t.next
t.next = node
node.next = self.head
self.tail = node
def delete(self, index):
if self.isEmpty():
print("The linked list is empty")
else:
t = self.head
l = self.length()
if index == 0:
self.head = t.next
self.tail.next = self.head
elif index == l - 1:
n = 1
while n < l - 1:
t = t.next
n += 1
t.next = self.head
self.tail = t
elif index > l - 1:
print("Out of range")
elif index < 0:
print("Wrong operation")
else:
n = 1
while n < index:
t = t.next
n += 1
a = t.next.next
t.next = a
def insert(self, data, index):
l = self.length()
if index == 0 or self.isEmpty():
self.addhead(data)
elif index >= l:
self.addtail(data)
else:
node = Node(data)
t = self.head
n = 1
while n < index:
t = t.next
n += 1
a = t.next
t.next = node
node.next = a
def search(self, a):
t = self.head
for i in range(a):
t = t.next
return t.data
def form(self, datalist):
self.addhead(datalist[0])
for i in range(1, len(datalist)):
self.addtail(datalist[i])
t = self.head
while t.next != self.head:
t = t.next
n, p = map(int, input().split(' '))
data = [i for i in range(1, n+1)]
print(data)
datalist = []
for i in range(len(data)):
datalist.append(data[i])
link = LinkList()
link.form(datalist)
a = p-1
b = []
while link.length() > 0:
b.append(link.search(a))
link.delete(a)
a += p-1
while a >= link.length():
a -= link.length()
print(b)
```
def append(self, data): new_node = Node(data) if not self.head: self.head = new_node self.tail = new_node else: new_node.prev = self.tail self.tail.next = new_node self.tail = new_node self.tail.next = self.head self.head.prev = self.tail一句一句解释这段代码
这段代码实现了在循环双向列表的尾部添加新节点的功能,下面是每一句代码的解释:
```python
def append(self, data):
# 创建一个新的节点
new_node = Node(data)
# 如果链表为空,则将新节点设置为头节点和尾节点
if not self.head:
self.head = new_node
self.tail = new_node
# 如果链表不为空,则将新节点添加到尾部
else:
# 将新节点的前驱指针指向原来的尾节点
new_node.prev = self.tail
# 将原来的尾节点的后继指针指向新节点
self.tail.next = new_node
# 将新节点设置为新的尾节点
self.tail = new_node
# 将尾节点的后继指针指向头节点,将头节点的前驱指针指向尾节点,以实现双向循环
self.tail.next = self.head
self.head.prev = self.tail
```
首先,在 `append` 方法内部,我们创建了一个新的节点 `new_node`,它包含要添加到列表中的数据。然后,在 `if not self.head` 的条件下,我们检查链表是否为空。如果是空的,那么将新节点设置为头节点和尾节点,因为在空链表中添加第一个节点时,它同时也是头节点和尾节点。
如果链表不为空,则我们将新节点添加到尾部。首先,我们将新节点的前驱指针指向原来的尾节点 `self.tail`,以便新节点可以链接到前一个节点。然后,我们将原来的尾节点的后继指针指向新节点 `new_node`,以便原来的尾节点可以链接到新节点。最后,我们将新节点设置为新的尾节点 `self.tail`。
最后,我们需要将尾节点的后继指针指向头节点 `self.head`,将头节点的前驱指针指向尾节点 `self.tail`,以实现双向循环。这样,在遍历列表时,我们可以通过尾节点的后继指针回到头节点,也可以通过头节点的前驱指针回到尾节点。