Your problem is handled in int remove_end(LinkedList *list). I've done the same as you and struggled at the same point, but finally implemented it cleanly. This change is not passed back to the caller. the question is specifically about the situation when there's only one element left. The headNode newHead assignment in popHead is assigning to the local headNode variable. Each node holds two items: The data (the actual content) A reference to the next node. I suspect that the word "last" in your question does not refer to the tail element of the list, but rather refers to the last remaining element of the list. Typically, each element of a linked list is called a node. The fact that you mention that head "is not pointer-to-pointer" suggests that this might indeed be important. I don't know whether that issue point is important in your case. For example, the caller's head pointer can be made accessible and modifiable from inside the delete function if the first parameter is declared as a pointer-to-pointer to head node int delete(struct node **phead, int item)Īlternatively delete function can be made to always return the updated head pointer value struct node *delete(struct node *head, int item) Typically, when one implements a function like that, one should make sure that the caller will know when the list becomes empty. It will point to now-deallocated memory, i.e. After such deletion the caller's head pointer will become invalid. However, if the list initially contained only one element (meaning that head is already pointing to the last element) then, of course, you can still easily delete it, but unfortunately you can't update caller's head pointer from inside delete function. When you add a node, set its next pointer to the old head, and it now becomes the new head of the list. Explain whats wrong with it and show how to fix the bug. Note that in that case the caller's head pointer will remain a perfectly valid pointer to a valid list. The following loop is supposed to delete all nodes from a linked list and release the memory they occupy. Simply iterate to the element before the last, delete the last and update the next pointer in new last element. Next contains a reference to the next node on the list. Data contains the value to be stored in the node. Each element of a linked list is called a node, and every node has two different fields. You can, of course, easily and safely delete the last element (= tail element of the list), if the list contains more than one element. Before going more in depth on what linked lists are and how you can use them, you should first learn how they are structured. I couldn’t solve it, so I looked at an answer that worked. Return the head of the merged linked list. This function takes a context value and returns a React node. Given a linked list, the task is to insert a new node at the end of the linked list. The list should be made by splicing together the nodes of the first two lists. 19 hours ago Thanks for contributing an answer to Stack Overflow. Merge the two lists in a one sorted list. The answer depends on what exactly is meant by the question. You are given the heads of two sorted linked lists list1 and list2.
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