leetcode284 : Peeking Iterator

1、原题如下：
iven 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.

2、解题如下：

// Below is the interface for Iterator, which is already defined for you.// **DO NOT** modify the interface for Iterator.class Iterator {    struct Data;    Data* data;public:    Iterator(const vector<int>& nums);    Iterator(const Iterator& iter);    virtual ~Iterator();    // Returns the next element in the iteration.    int next();    // Returns true if the iteration has more elements.    bool hasNext() const;};class PeekingIterator : public Iterator {public:    PeekingIterator(const vector<int>& nums) : Iterator(nums) {        // Initialize any member here.        // **DO NOT** save a copy of nums and manipulate it directly.        // You should only use the Iterator interface methods.    }    // Returns the next element in the iteration without advancing the iterator.    int peek() {        if(hasNext())        {            Iterator same(*this);                return same.next();        }    }    // hasNext() and next() should behave the same as in the Iterator interface.    // Override them if needed.    int next() {        return Iterator::next();    }    bool hasNext() const {        return Iterator::hasNext();    }};

3、总结
此题考查的利用迭代器原有的函数去定义构造新函数，其实peek–偷窥函数看起来很简单，但是由于迭代器在返回当前值的后就必须跳转到下一个值，所以我们这里采用的是用该迭代器的指针构造一个新的迭代器，让这个迭代器返回值，这样就避免了在原迭代器上操作的问题。当然也可以采用其他的方式求解。

4、PS
iterator中
next(), 是返回当前元素， 并指向下一个元素。
hasNext()， 则是判断当前元素是否存在，并指向下一个元素（即所谓的索引）