Peeking Iterator

题目地址：
https://leetcode.com/problems/peeking-iterator/description/

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?

解题思路：
这道题可以利用iterator的特性直接做。用next直接拿到下一个数，然后每次更新这个数就可以了。

代码：

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 rst = next;
        next = iter.hasNext() ? iter.next() : null;
        return rst;
    }

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

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