# All You Need to Know About Searching in a Linked List

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## The Algorithm for the Search in a Linked List

This algorithm can be implemented using the recursive and non-recursive way. Recursive solutions are generally more understandable and less optimal. For example, the recursive implementation of this task is almost two times shorter than the non-recursive, but it takes O(n) space, where n is the number of elements of a linked list.

When solving this problem, remember that you can choose the value of k so that when transferring k = 1 we obtain the last element, 2 – the penultimate, etc. Or choose k so that k = 0 corresponds to the last element.

Solution 1: linked list of known size

If the size of the linked list is known, the k-th element can be easily found from the end (L – k). All you need to do is go through the list and find the item.

Solution 2: recursive solution

This algorithm recursively iterates through a linked list. When the last element is reached, the algorithm starts the countdown, and the counter is reset to 0. Each step increments the counter by 1. When the counter reaches k, the desired item is found.

The implementation of this algorithm is quite short and simple: simply pass back an integer value through the stack. Unfortunately, the return operator can not return the value of the node. So how do we get around this difficulty?

Approach A: do not return the element

We can forget about the return of the item, as it is enough to withdraw it as soon as it is found. And in the “return” operator we can return the value of the counter.

public static int counter(node head, int k) {

return 0;

}

int i = counter(head.next, k) + 1;

if (i == k) {

}

return i;

}

Solution is correct, but we can go the other way.

Approach B: use C++

The second method is the use of C++ and transferring the value by reference. This approach allows us not only to return the value of the node but also to update the counter by passing a pointer to it.

node* counter(node* head, int k, int& a) {

return NULL;

}

node* nod = counter(head->next, k, a);

a = a + 1;

if (a == k) {

}

return nod;

}

Solution 3: iterative solution

The iterative solution is more complex, but also more optimal. We can use two pointers: pointer1 and pointer2. Initially, both pointers point to the top of the list. Then we move forward pointer2 by k nodes. Now we are beginning to move both pointers at the same time. When pointer2 reaches the end of the list, pointer1 will point to the item we want it to.

node counter(node head, int k) {

if (k <= 0) return 0;

for (int i = 0; i < k – 1; i++) {

if (pointer2 == null) return null;

pointer2 = pointer2.next;

}

if (pointer2 == null) return null;

while (pointer2.next != null) {

pointer1 = pointer1.next;

pointer2 = pointer2.next;

}

return pointer1;

}