[LeetCode] Peeking Iterator

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.

Hint:

Think of “looking ahead”. You want to cache the next element. Is one variable sufficient? Why or why not? Test your design with call order of peek() before next() vs next() before peek(). For a clean implementation, check out Google’s guava library source code.

解題思路

用一個變量來保存peek元素值，每次調用next()和peek()時更新其值。

實現代碼

Java：

// Runtime: 120 ms
//Java Iterator interface reference:
//https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html
class PeekingIterator implements Iterator {
    Integer peek = null;
    private Iterator iterator;
    public PeekingIterator(Iterator iterator) {
        this.iterator = iterator;
    }

 // Returns the next element in the iteration without advancing the iterator.
    public Integer peek() {
        if (peek == null) {
            peek = iterator.next();
        }
        return peek;
    }

    // hasNext() and next() should behave the same as in the Iterator interface.
    // Override them if needed.
    @Override
    public Integer next() {
        if (peek == null) {
            return iterator.next();
        }
        else {
            Integer temp = peek;
            peek = null;
            return temp;
        }
    }

    @Override
    public boolean hasNext() {
        if (peek != null) {
            return true;
        }
        else {
            return iterator.hasNext();
        }
    }
}