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().

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:

1. Think of “looking ahead”. You want to cache the next element.
2. Is one variable sufficient? Why or why not?
3. Test your design with call order of peek() before next() vs next() before peek().
4. For a clean implementation, check out Google’s guava library source code.

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

这道题让我们实现一个顶端迭代器，在普通的迭代器类Iterator的基础上增加了peek的功能，就是返回查看下一个值的功能，但是不移动指针，next()函数才会移动指针，那我们可以定义一个变量专门来保存下一个值，再用一个bool型变量标记是否保存了下一个值，再调用原来的一些成员函数，就可以实现这个顶端迭代器了，参见代码如下：

解法一：

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) {
        _flag = false;
    }

    int peek() {
        if (!_flag) _value = Iterator::next();
        _flag = true;
        return _value;
    }

    int next() {
        if (!_flag) return Iterator::next();
        _flag = false;
        return _value;
    }

    bool hasNext() const {
        return _flag || Iterator::hasNext();
    }

private:
    int _value;
    bool _flag;
};

这道题主要的考点就是peek函数，因为这个是默认的迭代器不具备的功能。我们其实可以使用一个小trick来实现peek功能，由于peek是要暗中观察一下下一个元素，但是迭代器并不真正移动到下一个，那么我们其实是可以创建一个副本，然后让副本移动到下一个，并返回，由于是局部变量，副本在调用结束后也会被销毁，所以并没有任何内存问题，可以说是一种相当聪明的解法了，参见代码如下：

解法二：

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) {}

    int peek() {
        return Iterator(*this).next();
    }

    int next() {
        return Iterator::next();
    }

    bool hasNext() const {
        return Iterator::hasNext();
    }
};

类似题目：

Binary Search Tree Iterator

Flatten 2D Vector

Zigzag Iterator

参考资料：

https://leetcode.com/problems/peeking-iterator/

https://leetcode.com/problems/peeking-iterator/discuss/72650/10-line-C%2B%2B-and-14-line-Java-Implementation

https://leetcode.com/problems/peeking-iterator/discuss/72554/Simple-C%2B%2B-solution-(1-line-per-method)-without-extra-member-variables