[LeetCode]237. Delete Node in a Linked List(删除链表结点)

237. 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.

编写一个函数来删除单链表中的结点（尾部除外），只允许访问该结点。

Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -> 4 after calling your function.

假设链表是1 - > 2 - > 3 - > 4，你被赋予值为3的第三个结点，在调用函数后，链表应该变为1 - > 2 - > 4。

思路：

删除单链表的某个结点，只需把后一个结点挂到前一个结点后面就行

正好利用了上学期学的数据结构知识。

代码如下：

C++

#include <iostream>

using namespace std;

//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 = *node->next; //直接把后一个结点赋给现在的节结点,但没有彻底删除数据，只是把结点从链表上去掉

auto next = node->next;
*node = *next;
delete next;//彻底删除结点
}
};
int main()
{
Solution a;
cout << "Hello world!" << endl;
return 0;
}

C# or Java

public void deleteNode(ListNode node) {
node.val = node.next.val;
node.next = node.next.next;
}