【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++   /*刪除連結串列中的某一個結點（本題不用考慮尾結點） 通用方法（給的是連結串列頭結點）：可以從頭結點開始遍歷到要刪除結點的上一個結點，然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即可（O（1）），         但是對尾結點，則只能從連結串列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; //*為取內容運算子，將指標指向的結構體內容複製過去                 //真的刪除       /*  ListNode* temp = node->next;         *node = *temp; //將指標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 ;  //儲存下一個結點指標，以便之後刪除     node -> val = pnext -> val ;   //複製值     node -> next = pnext -> next ;  //複製指標     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;   */