leetcode--237:(链表类) Delete Node in a Linked List

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Given linked list -- head = [4,5,1,9], which looks like following:


Note:

  • The linked list will have at least two elements.
  • All of the nodes' values will be unique.
  • The given node will not be the tail and it will always be a valid node of the linked list.
  • Do not return anything from your function.

我的思路:

  这是个单向链表,无法得知node的上一个节点是谁;

  node的val用node.next.val代替,然后node.next指向node.next.next,这样即使是单向链表,删除节点还是不会断。

代码实现:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

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

 

原文地址:https://www.cnblogs.com/marvintang1001/p/11171779.html