如何将链表从某个键反转到末尾?(蟒蛇3)

Question

我试图在python中构造一个方法，以反转从某个键开始到最后的单链表。我知道如何反转整个链表，但我不会编码

def reverse(self, key)

• 键不是节点，而是值

Answer 1

答案是首先初始化一个双链接列表，但我们将在这个场景中模拟它的用例，并创建一个临时的"先前的"；节点。

class LinkedList:
def __init__(self):
self.head = None
self.tail = None
def insert(self, key):
node = Node(key)
if self.tail:
self.tail.next = node
self.tail = node
else:
self.tail = self.head = node
def reverse(self, key):
first = []
prev = None
current, self.head, self.tail = self.head, self.tail, self.head
while current:
if current.value == key:
first.append(current)
current.next, current, prev = prev, current.next, current
return first
def list_nodes(self):
res = []
current = self.head
while current:
res.append(current.value)
current = current.next
return res

class Node:
def __init__(self, value):
self.next = None
self.value = value

if __name__ == '__main__':
lst = LinkedList()
for i in range(100):
lst.insert(i)
res1 = lst.list_nodes()
rev = lst.reverse(55)
res2 = lst.list_nodes()

输出为：

res1 = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99]
res2 = [99, 98, 97, 96, 95, 94, 93, 92, 91, 90, 89, 88, 87, 86, 85, 84, 83, 82, 81, 80, 79, 78, 77, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
rev = [<__main__.Node object at 0x00000271B63241C0>]  # A Node object at some location

这将反转并搜索该键，并在列表中返回它，最小的索引是最接近的。