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.

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:

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.

Solution:

刚开始的思路是, 用被删除结点前的前一个结点prev, prev->next = node->next; 再释放node的内存。

但因为是单向链表, 只给出一个结点, 因此要删除这个结点就不能找prev结点.

将下一个结点的值赋给该结点, 该结点指向跳过下一结点, 释放下一个结点。实际上是删除了下一个结点, 但将值赋值给了当前node

/**
 * 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;
      ListNode* delNode = node->next;
      node->next = delNode->next;
      delete(delNode);
    }
};