[LeetCode] 284. Peeking Iterator Java

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().

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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?

题意及分析：迭代器Iterator 接口有next和hasnext方法，PeekingIterator 实现了Iterator 接口，要求实现一个peek方法，返回下一个元素，但是不移动到下一个元素。这里初始化的时候便向后移动一位，然后在next()中使用一个中间变量res保存next()的值，然后返回res即可。

代码：

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

    private Integer next = null;
    private Iterator<Integer> iter;
    public PeekingIterator(Iterator<Integer> iterator) {
        // initialize any member here.
        iter = iterator;
        if (iter.hasNext())
            next = iter.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 = iter.hasNext() ? iter.next() : null;
        return res;
    }

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