[Leetcode] Peeking Iterator 瞥一眼迭代器

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().Show More Hint Follow up: How would you extend your design to be generic and work with all types, not just integer?

Credits: Special thanks to @porker2008 for adding this problem and creating all test cases.

缓存法

复杂度

时间 O(1) 空间 O(1)

思路

为了能peek后下次next还得到同样的数字，我们要用一个缓存保存下一个数字。这样当peek时候，返回缓存就行了，迭代器位置也不会变。当next的时候除了要返回缓存，还要将缓存更新为下一个数字，如果没有下一个就将缓存更新为null。

代码

class PeekingIterator<T> implements Iterator<T> {

    T cache;
    Iterator<T> it;

    public PeekingIterator(Iterator<T> iterator) {
        // initialize any member here.
        this.cache = iterator.next();
        this.it = iterator;
    }

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

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

    @Override
    public boolean hasNext() {
        return it.hasNext() || cache != null;
    }
}

后续 Follow Up

Q：如何支持任意类型的迭代器？
A：使用Generic的方式，代码中已经将Integer替换为Generic的T