My code:

// Java Iterator interface reference:
// https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html
class PeekingIterator implements Iterator<Integer> {
    private Iterator<Integer> iterator;
    private Integer pre;
    public PeekingIterator(Iterator<Integer> iterator) {
        // initialize any member here.
        this.iterator = iterator;
        if (this.iterator.hasNext()) {
            pre = this.iterator.next();
        }
        else
            pre = null;
    }

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

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

    @Override
    public boolean hasNext() {
        return pre != null;
    }
}

这道题目没什么难度。就是pre 指向当前 next, next往后移动一格。
然后当hasNext 返回false时，pre = null
判断 hasNext, 也特别简单。只用判断 pre == null

Follow up 说能不能把Integer 改成其他任意类型。我的做法是可以的。

Anyway, Good luck, Richardo!

My code:

// Java Iterator interface reference:
// https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html
class PeekingIterator implements Iterator<Integer> {
    Integer pre = null;
    Iterator<Integer> iter;
    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 (pre == null) {
            pre = iter.next();
        }
        return pre;
    }

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

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

题目并不难。也可以改成泛型。
reference:
https://github.com/google/guava/blob/703ef758b8621cfbab16814f01ddcc5324bdea33/guava-gwt/src-super/com/google/common/collect/super/com/google/common/collect/Iterators.java#L1125

Anyway, Good luck, Richardo! -- 09/12/2016