【LeetCode & 剑指offer刷题】链表题3：18 删除链表中的结点（237. Delete Node in a Linked List）

【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）

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.

---

C++

 

/*删除链表中的某一个结点（本题不用考虑尾结点）

通用方法（给的是链表头结点）：可以从头结点开始遍历到要删除结点的上一个结点，然后该结点指向要删除结点的下一个结点，删除要删除的结点，不过需花费O（n）

方法2(给的是要删除结点)：对于非尾结点，将下个结点的内容复制到本结点，在删除掉下一个结点即可（O（1）），

        但是对尾结点，则只能从链表头结点开始遍历到尾结点的前一个结点(O(n))

*/

/**

 * 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; //*为取内容运算符，将指针指向的结构体内容复制过去

       

        //真的删除

      /*  ListNode* temp = node->next;

        *node = *temp; //将指针指向的结构体内容复制过去 

        delete temp; //删除多余的结点*/

       

    if(node == nullptr) return;

    ListNode* pnext = node->next; //保存下一个结点指针，以便之后删除

    node->val = pnext->val;  //复制值

    node->next = pnext->next; //复制指针

    delete pnext; //掌握

       

    }

};

/*

也可用这个（掌握）

    ListNode* pnext = node->next;

    node->val = node->next->val;

    node->next = node->next->next;

    delete pnext;

  */