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.
删除除了尾部外的链表中的值,一开始的做法没有考虑到释放内存,只是简单的移动赋值,奇怪的居然也AC了。
1 void deleteNode(struct ListNode* node) { 2 if(node){ 3 node->val=node->next->val; 4 node->next=node->next->next; 5 } 6 }
在讨论区看到别人更好的做法
1 void deleteNode(struct ListNode* node) { 2 if(node){ 3 struct ListNode *next=node->next; 4 *node=*next; 5 free(next); 6 } 7 }