Linked List: Searching

Task: Find 30 in this list

class Node { int value; Node next; } Node n = head;              // start at the headwhile (n != null) {         // walk to the end    n = n.next;             // follow one pointer} n.next = prev.next;         // insert: point it onprev.next = n;              // then point prev at itprev.next = prev.next.next; // delete: step over one

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Searching a linked list is traversal that stops early. Each node is compared with the target and, if it does not match, the next pointer is followed. 30 sits third, so it costs three steps to reach - and no arrangement of the values would help, because a linked list cannot jump to the middle to begin a binary search.