【LeetCode & 劍指offer刷題】鏈表題3：18 刪除鏈表中的結點（237. Delete Node in a Linked List）

【LeetCode & 劍指offer 刷題筆記】目錄（持續更新中...）

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++

 

/*刪除鏈表中的某一個結點（本題不用考慮尾結點）

通用方法（給的是鏈表頭結點）：能夠從頭結點開始遍歷到要刪除結點的上一個結點，而後該結點指向要刪除結點的下一個結點，刪除要刪除的結點，不過需花費O（n）

方法2(給的是要刪除結點)：對於非尾結點，將下個結點的內容複製到本結點，在刪除掉下一個結點便可（O（1）），

        可是對尾結點，則只能從鏈表頭結點開始遍歷到尾結點的前一個結點(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; //*爲取內容運算符，將指針指向的結構體內容複製過去

       

        //真的刪除

      /*  ListNode* temp = node->next;

        *node = *temp; //將指針指向的結構體內容複製過去 

        delete temp; //刪除多餘的結點*/

       

    if ( node == nullptr ) return ;

    ListNode * pnext = node -> next ;  //保存下一個結點指針，以便以後刪除

    node -> val = pnext -> val ;   //複製值

    node -> next = pnext -> next ;  //複製指針

    delete pnext ; //掌握

       

    }

};

/*

也可用這個（掌握）

    ListNode* pnext = node->next;

    node->val = node->next->val;

    node->next = node->next->next;

    delete pnext;

  */