[LeetCode 284] Peeking Iterator-优快云博客

本文详细解析了PeekingIterator的设计思路与实现方法，通过继承Iterator接口并扩展peek功能，实现了在不移动迭代器的情况下预览下一个元素的能力。文章通过具体代码示例展示了如何在C++中实现这一功能，同时讨论了如何使设计通用化以适应不同数据类型。

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 callpeek()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?

分析

这道题目看起挺复杂的，其实需要仔细的分析一下，发现其实并不难。题目中已经给我们提供了基类的方法和提供的函数。我们需要利用基类Iterator的函数实现子类的peek。如果我们想知道peek的值，其实我们只要有保存一个Iterator，比当前的Iterator往后移一位，并记录值，这个值就是peek的值。当调用next时，只需要同样调用next并记录值就行了。

Code

// 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.
        iter = new Iterator(nums);
        peekNum = iter->next();
	}

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

	// hasNext() and next() should behave the same as in the Iterator interface.
	// Override them if needed.
	int next() {
        if (iter->hasNext())
            peekNum = iter->next();
        return Iterator::next();
	}

	bool hasNext() const {
        return Iterator::hasNext();
	}
    
    Iterator* iter;
    int peekNum;
};

运行效率

Runtime: 8 ms, faster than 79.95% of C++ online submissions for Peeking Iterator.

Memory Usage: 9.9 MB, less than 59.46% of C++ online submissions for Peeking Iterator.