Peeking Iterator

Discussion

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.

Solution

// 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.
        has_peeked = false;
    }

    // Returns the next element in the iteration without advancing the iterator.
    int peek() {
        if(has_peeked == false) {
            has_peeked = true;
            value = Iterator::next();
        }
        return value;
    }

    // hasNext() and next() should behave the same as in the Iterator interface.
    // Override them if needed.
    int next() {
        //has_peeked is true means "value" has been updated in peek()
        if(has_peeked == false) return Iterator::next();
        has_peeked = false;//now the current value is not peeked yet
        return value;
    }

    bool hasNext() const {
        if(has_peeked) return true;
        return Iterator::hasNext();
    }
private:
    bool has_peeked = false;
    int value;
};

还有个解法是利用Iterator 的copy constructor.
Iterator(*this) makes a copy of current iterator, then call next on the copied iterator to get the next value without affecting current iterator.

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();
    }
};

Reference

http://www.cnblogs.com/grandyang/p/4825068.html