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LeetCode#237. Delete nodes in the linked list
2022-07-06 15:21:00 【Rufeng ZHHH】
Please write a function , be used for Delete a specific node in the single linked list . When designing functions, you need to pay attention to , You can't access the head node of the linked list head , Direct access only The node to be deleted .
The topic data ensures that the nodes to be deleted Not the end node .
Example 1:
Input :head = [4,5,1,9], node = 5
Output :[4,1,9]
explain : Specifies that the value in the linked list is 5 Second node of , So after calling your function , The list should be 4 -> 1 -> 9
Example 2:
Input :head = [4,5,1,9], node = 1
Output :[4,5,9]
explain : Specifies that the value in the linked list is 1 The third node of , So after calling your function , The list should be 4 -> 5 -> 9
Example 3:
Input :head = [1,2,3,4], node = 3
Output :[1,2,4]
Example 4:
Input :head = [0,1], node = 0
Output :[1]
Example 5:
Input :head = [-3,5,-99], node = -3
Output :[5,-99]
Tips :
The number range of nodes in the linked list is [2, 1000]
-1000 <= Node.val <= 1000
The value of each node in the linked list is unique
Nodes to be deleted node yes A valid node in the linked list , And Not the end node
source : Power button (LeetCode)
link : Power button
This problem can be solved as long as we delete the specified node , Because the title has stated that this node is not the tail node , We can put node.val Locate the node.next.val, meanwhile node.next=node.next.next that will do .
# Definition for singly-linked list.
# 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
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