Data Structures and Algorithm Analysis - Computer Science at ...

Sec. 4.1 Lists 115/** Insert "it" at current position */public void insert(E it) {curr.setNext(new DLink(it, curr, curr.next()));curr.next().next().setPrev(curr.next());cnt++;}/** Append "it" to list */public void append(E it) {tail.setPrev(new DLink(it, tail.prev(), tail));tail.prev().prev().setNext(tail.prev());cnt++;}/** Remove and return current element */public E remove() {if (curr.next() == tail) return null; // Nothing to removeE it = curr.next().element(); // Remember valuecurr.next().next().setPrev(curr);curr.setNext(curr.next().next()); // Remove from listcnt--;// Decrement the countreturn it;// Return value removed}/** Move curr one step left; no change if at front */public void prev() {if (curr != head) // Can’t back up from list headcurr = curr.prev();}Figure 4.15 Implementations for doubly linked list insert, append,remove, and prev methods.Example 4.1 There is a space-saving technique that can be employed toeliminate the additional space requirement, though it will complicate theimplementation and be somewhat slower. Thus, this is an example of aspace/time tradeoff. It is based on observing that, if we store the sum oftwo values, then we can get either value back by subtracting the other. Thatis, if we store a + b in variable c, then b = c − a and a = c − b. Of course,to recover one of the values out of the stored summation, the other valuemust be supplied. A pointer to the first node in the list, along with the valueof one of its two link fields, will allow access to all of the remaining nodesof the list in order. This is because the pointer to the node must be the sameas the value of the following node’s prev pointer, as well as the previousnode’s next pointer. It is possible to move down the list breaking apartthe summed link fields as though you were opening a zipper. Details forimplementing this variation are left as an exercise.