LeetCode: 284. Peeking Iterator

LeetCode: 284. Peeking Iterator

题目描述

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

解题思路

将当前迭代器记录的数据暂存起来，并让迭代器指向下一个元素。每次 ​​peek()​​ 时，返回暂存的元素。

AC 代码

// 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 {
private:
    int m_curNum;
    bool m_hasNext;
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.
        if(Iterator::hasNext())
        {
            m_curNum = Iterator::next();
            m_hasNext = true;
        }
        else
        {
            m_hasNext = false;
        }
    }

    // Returns the next element in the iteration without advancing the iterator.
    int peek() {
        return m_curNum;
    }

    // hasNext() and next() should behave the same as in the Iterator interface.
    // Override them if needed.
    int next() {
        int lastNum = m_curNum;
        
        if(Iterator::hasNext())
        {
            m_curNum = Iterator::next();
            m_hasNext = true;
        }
        else
        {
            m_hasNext = false;
        }
        
        return lastNum;
    }

    bool hasNext() const {
        return m_hasNext;
    }
};