LeetCode 237. Delete Node in a Linked List（链表题目）

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:

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:

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.

思路分析

该题主要是删除一个节点，但是不知道该节点的前一个节点，就没有办法直接将该节点删除之后，使前一个结点指向删除结点的后一个，所以传统的删除方法不能用。但是，我们可以考虑删除节点的后一个节点，将其值赋给删除结点，然后删除后一个节点，建立起新的指针指向关系即可。

具体代码：

/**
 * 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) {
        
        if(node==NULL)
            return ;
        if(node->next==NULL)
        {
            delete node;
            node=NULL;
        }
        
        //采用删除下一个节点的思想
        node->val=node->next->val;
        ListNode* del=node->next;
        node->next=del->next;
        
        delete del;

    }
};