284. Peeking Iterator

// Java Iterator interface reference:
// https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html

class PeekingIterator implements Iterator<Integer> {
    private Integer top;
    private Iterator<Integer> iterator;

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

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

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

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

把这个抽象为给枪上子弹. 枪管里只能有一颗子弹, 弹匣里可以有很多. peek()就是看枪膛里的子弹是什么, next()则是看枪膛的子弹是什么, 发射这个子弹并且再次填装. hasNext()是看枪膛中有没有子弹.

这里的top就表示枪膛里的子弹. next()的话就是首先记录枪膛的子弹是什么, 然后看弹匣中还有子弹吗(iterator.hasNext()), 如果有那就让top等于iterator.next(), 没有就让top等于null. 然后我们通过判断top的值来确实hasNext()返回什么.

时间复杂度和空间复杂度都是O(1).