题目地址：https://leetcode.com/problems/delete-node-in-a-linked-list/open in new window
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.
Given linked list -- head = [4,5,1,9], which looks like following:
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、 Thelinkedlistwillhaveatleasttwoelements.；
2、 Allofthenodes'valueswillbeunique.；
3、 Thegivennodewillnotbethetailanditwillalwaysbeavalidnodeofthelinkedlist.；
4、 Donotreturnanythingfromyourfunction.；
题目大意
给出了一个节点，这个节点是在一个单链表中的，并且这个节点不是最后一个节点。现在要我们删除这个节点。
解题方法
设置当前节点的值为下一个
拿到这个题时以为要从头找到这个节点前一个节点，然后删除当前节点。这个应该是正常思路。
但是，题目只给出了删除的这个节点，没有给出根节点。所以可以通过这个将当前的节点的数值改成下面的节点的值，然后删除下一个节点的方式。
链表基本操作，记待删除节点为node：
令node.val = node.next.val，node.next = node.next.next即可
Java代码如下：
/**
* Definition for singly-linked list.
* 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;
}
}
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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
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C++代码如下：
/**
* Definition for singly-linked list.
* 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;
}
};
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