Open huimeich opened 8 years ago
Hint Think of "looking ahead". You want to cache the next element
// Java Iterator interface reference:
// https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html
class PeekingIterator implements Iterator<Integer> {
private Iterator<Integer> iter;
private Integer top;
public PeekingIterator(Iterator<Integer> iterator) {
// initialize any member here.
this.iter = iterator;
}
// Returns the next element in the iteration without advancing the iterator.
public Integer peek() {
if (top != null) return top;
top = iter.next();
return top;
}
// hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
@Override
public Integer next() {
if (top != null) {
Integer temp = top;
top = null;
return temp;
} else {
return iter.next();
}
}
@Override
public boolean hasNext() {
if (top == null && !iter.hasNext()) return false;
return true;
}
}
Given an Iterator class interface with methods: next() and hasNext(), design and implement a PeekingIterator that support the peek() operation -- it essentially peek() at the element that will be returned by the next call to next().
Here is an example. Assume that the iterator is initialized to the beginning of the list: [1, 2, 3].
Call next() gets you 1, the first element in the list.
Now you call peek() and it returns 2, the next element. Calling next() after that still return 2.
You call next() the final time and it returns 3, the last element. Calling hasNext() after that should return false.