# 237. Delete Node in a Linked List 删除链表中的节点

@TOC

[LeetCode]

Total Accepted: 78258 Total Submissions: 179086 Difficulty: Easy

## # 题目描述

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

``````4 -> 5 -> 1 -> 9
``````

Example 1:

``````Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
Explanation: You are given the second node with value 5, the linked list
should become 4 -> 1 -> 9 after calling your function.
``````

Example 2:

``````Input: head = [4,5,1,9], node = 1
Output: [4,5,9]
Explanation: You are given the third node with value 1, the linked list
should become 4 -> 5 -> 9 after calling your function.
``````

Note:

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

## # 解题方法

### # 设置当前节点的值为下一个

Java代码如下：

``````	/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public void deleteNode(ListNode node) {
node.val=node.next.val;
node.next=node.next.next;
}
}
``````

python代码如下：

``````# 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.
"""
node.val = node.next.val
node.next = node.next.next
``````

C++代码如下：

``````/**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void deleteNode(ListNode* node) {
node->val = node->next->val;
node->next = node->next->next;
}
};
``````

## # 日期

2016/4/30 0:39:40 2018 年 11 月 11 日 —— 剁手节快乐 2019 年 9 月 27 日 —— 昨天面快手，竟然是纯刷题