Insert newNode into a linked list after targetNode
Node currentNode = root
while currentNode != targetNode
currentNode = currentNode.next
newNode.next = currentNode.next
currentNode.next = newNode
The time complexity of inserting an element into a linked list is O(1) or constant time.
The time complexity to find an element in a linked list is O(n), where n is the number of elements in the list. This means that the time it takes to find an element in a linked list increases linearly with the number of elements in the list.
To efficiently sort a doubly linked list, you can use a sorting algorithm such as merge sort or quicksort. These algorithms can be implemented to work with doubly linked lists by considering the pointers in both directions. By recursively dividing the list and merging or partitioning the elements, you can achieve an efficient sorting process.
The primary advantage of a binary search algorithm is that searching a sequence can be achieved in logarithmic time. The primary disadvantages are that the data must be in sorted order. Arrays are the ideal container for binary searches as they provide constant-time random-access and are therefore trivial to both sort and search efficiently. Sorted binary trees can also be used, however for optimal performance they must be totally balanced (e.g., red/black binary tree). Constant-time random-access is not a requirement of binary trees, however the cost of maintaining balance during construction of the tree has to be taken into account. With a linear search, we start at one end of the sequence and traverse through the sequence one element at a time until we find the value we're looking for, or we reach the element one-past-the-end of the sequence (in which case the element we're looking for does not exist). For a sequence of n elements, the worst case is O(n). Linear search is ideal for forward lists (singly-linked lists) and lists (doubly-linked lists) as neither provides nor requires constant-time random-access. With binary search, we locate the middle element in the sequence. If that's not the value we are looking for, we can easily determine which half of the sequence contains our value because the elements are in sorted order. So we eliminate the other half and repeat the algorithm with the remaining half. As such, each failure to find the value reduces the number of elements to be searched by half (plus the middle element). If there are no elements in the remaining half then the value does not exist. The worst case is therefore O(log n).
In a linked list data structure, the head is the starting point that points to the first node in the list. It is significant because it allows for traversal of the list by providing access to the first element, enabling operations such as insertion, deletion, and searching.
theory part to how u can insert element at middle of link list ?
The time complexity of inserting an element into a linked list is O(1) or constant time.
Given a currentNode pointer, perform the following steps: Create a newNode. Assign currentNode->next to newNode->next. Assign currentNode to newNode->prev. Assign newNode to currentNode->next.
Linear time. O(n).
I am telling you the algorithm of inserting an element at any location of simply link list:-insert_loc(start,item,loc)step1. [Check for overflow]if ptr=NULLthen print overflowexitelse[ptr=(node*) malloc (size of nodes)] {memory allocation}step2. set ptr->info=itemstep3. set i=1set temp=startstep4. Repeat step 5&6 until inextstep6. set i=i+1step7. set ptr->next=temp->nextstep8. set temp->next=ptrGood luckRjames007
To delete a linked list walk through the list and delete the memory allocated to each element, remembering the next element address, and then iterating or recursing the process using the next element address, until the next element address is null.
Algorithm to insert an element at the end of a linked listSpecial case: the list is empty.Create a new node for the element, assigning nullptr (0) to its next node.Assign the new node to the head of the list.Return a pointer to the new node and exit.All other cases: the list is not empty.Start at the head node (make it current).Repeat: while the current node has a next node, traverse to that node (make it current).Create a new node for the element, assigning nullptr (0) to its next node.Assign the new node as the current node's next node.Return a pointer to the new node and exit.
To search for an element in a linked list, you iterate the list, looking for the element, and either return the element or an indication that it was not found. for (ptr = first; ptr != null; ptr = ptr.next) if (ptr.value == searchvalue) break; This will either leave ptr with the address of the found element, or null, if not found.
To insert an element in a linked list, you need a pointer to the new element, and a pointer to the target element. Set new.next = target.next, and then set target.next = new. This will insert new after target. To insert before target, you need the pointer to the element brfore target, and you can find that by searching from top until you find an element where element.next == target. To insert at top, a special case, you set new.next = top, and then set top = new.
pseudo code algorithm to create a linked list
sorry
Linked lists are best suited to dynamic collections where random access is not a major concern, such as queues and stacks. Using an array would prove highly inefficient in these cases due to the need to resize the array to accommodate new elements or to remove unused elements. Occasionally, an increase in size will require the entire array be copied to a larger block of free memory. Linked lists don't have this problem as memory is allocated and de-allocated on a per element basis.