LeetCode 284. Peeking Iterator

Description

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.

Framework

题目给出的框架如下，可以看出这个PeekingIterator继承了Iterator。

/*
 * 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;
 *		Iterator(const vector<int>& nums);
 * 		Iterator(const Iterator& iter);
 *
 * 		// 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() {
	}
	
	// hasNext() and next() should behave the same as in the Iterator interface.
	// Override them if needed.
	int next() {
	}
	
	bool hasNext() const {
	}
};

Idea

根据题目的意思来看，hasNext和next的方法和父类一致，直接调用父类的方法即可，而peek方法则需要在取数的同时不往后迭代。那么一个可行的思路就是，用this指针构造一个父类的对象，再调用其next方法。

Solution

最终代码如下：

```cpp
/*
 * 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;
 *		Iterator(const vector<int>& nums);
 * 		Iterator(const Iterator& iter);
 *
 * 		// 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() {
        return Iterator(*this).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();
	}
};

这道题并没有考察什么复杂的算法，而是考察了类的继承里的一些基本的概念，可以说是一道很有趣的题目了。