[leetcode]Peeking Iterator

题目描述如下：

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.

对Iterator接口的改造，说实话没有什么好讲的，附上代码：

class PeekingIterator implements Iterator<Integer> {

    Iterator<Integer> myIterator;
    Integer nextValue;

    public PeekingIterator(Iterator<Integer> iterator) {
        // initialize any member here.
        this.myIterator = iterator;
        if(iterator.hasNext()) nextValue = iterator.next();
    }

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

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

    @Override
    public boolean hasNext() {
        if(nextValue == null) return false;
        else return true;
    }
}

题目链接：https://leetcode.com/problems/peeking-iterator/