删除中间节点
实现一种算法，删除单向链表中间的某个节点（即不是第一个或最后一个节点），假定你只能访问该节点。
示例：
输入：单向链表a->b->c->d->e->f中的节点c 结果：不返回任何数据，但该链表变为a->b->d->e->f
示例代码：
# Definition for singly-linked list.
class LinkNode(object):
def __init__(self, x):
self.val = x
self.next = None
class Solution(object):
# create a linklist
def creat_linklist(self, li):
if not li:
print('This is an empty list')
return
head = p = LinkNode(li[0])
for i in li[1:]:
p.next = LinkNode(i)
p = p.next
return head
def print_linklist(self, head):
if not head:
print('This is an empty list')
return
while head:
print(head.val, end="->")
head = head.next
print('None')
def del_node(self, head, node):
if not head:
print('This is an empty list')
return
while head:
if head.val == node:
head.val = head.next.val
head.next = head.next.next
head = head.next
# test
li = ['a', 'b', 'c', 'd', 'e', 'f']
obj = Solution()
head = obj.creat_linklist(li)
obj.print_linklist(head)
obj.del_node(head, 'c')
obj.print_linklist(head)
运行结果：
