2017-11-26 16:48:09 +0000   |     algorithm leetcode design   |   Viewed times   |    

题目

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.

Follow up: How would you extend your design to be generic and work with all types, not just integer?

用一个变量缓存下一个要返回的数字

因为传进来的是一个经典的迭代器,所以最朴素的做法,如果这个迭代器的输出和我们要的输出差别较大,且不可控,最差的情况我们可以预先把迭代器里的内容全部读取到内存。

当然这题没必要这么做。因为可以直接利用原始迭代器的输出。唯一需要处理的问题是原始迭代器的next()方法获取到下一个元素之后,无法再次获取同一个元素。所以解决办法很简单,就是把下一个元素缓存起来。

代码

// Java Iterator interface reference:
// https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html
class PeekingIterator implements Iterator<Integer> {

    private Iterator<Integer> ite = null;
    private Integer next = null; // 缓存下一个要返回的元素

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

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

	// hasNext() and next() should behave the same as in the Iterator interface.
	// Override them if needed.
	@Override
	public Integer next() {
	    Integer res = next;
        next = (ite.hasNext())? ite.next() : null;
        return res;
	}

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

结果

peeking-iterator-1