leetcode 284. Peeking Iterator

秒速五厘米 2022-07-26 00:26 132阅读 0赞

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().

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Here is an 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.

// 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 {
    	vector<int>num;
    	int k = 0;
    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.
    		num = nums;
    	}
    	// Returns the next element in the iteration without advancing the iterator.
    	int peek() {
    		return num[k];
    	}
    	// hasNext() and next() should behave the same as in the Iterator interface.
    	// Override them if needed.
    	int next() {
    		return num[k++];
    	}
    	bool hasNext() const {
    		return k < num.size();
    	}
    };