每日一题之LeetCode237删除链表中的节点876链表的中间节点-优快云博客

本文介绍了一种链表节点删除的方法，并实现了一个在链表中查找中间节点的算法。通过具体实例展示了如何在链表中删除指定节点以及如何找到链表的中间节点。

237
class ListNode:
def init(self, x):
self.val = x
self.next = None

class Solution:
def deleteNode(self, node):
"""
:type node: ListNode
:rtype: void Do not return anything, modify node in-place instead.
"""

    node.val=node.next.val
    node.next=node.next.next

head1 = ListNode(4)
n1 = ListNode(5)
n2 = ListNode(1)
n3 = ListNode(9)
head1.next = n1
n1.next = n2
n2.next = n3
t=Solution()
t.deleteNode(n1)
print(head1.next.val)

876

-- coding: utf-8 --

"""
Created on Thu Nov 22 21:45:28 2018

@author: Administrator
"""

class ListNode:
def init(self, x):
self.val = x
self.next = None

class Solution(object):
def middleNode(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
flag=True
i=0
t=head
while flag:
if t.next!=None:
i=i+1
t=t.next
else:
flag=False
i=i+1
j=int(i/2)+1

while k:

head=head.next

n=n+1

if n=j:

k=False

    for m in range(0,j-1):

        head=head.next
    return head

head1 = ListNode(1)
n1 = ListNode(2)
n2 = ListNode(3)
n3 = ListNode(4)
n4 = ListNode(5)
head1.next = n1
n1.next = n2
n2.next = n3
n3.next = n4
t=Solution()
print(t.middleNode(head1).val)
#print(head1.next.val)

转载于:https://blog.51cto.com/13930723/2320836