> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/284.peeking_iterator.md).

# 284.Peeking Iterator

**284.Peeking Iterator**

难度:Medium

> 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?

We can set a right value stands for next value, a bool stands for hasNext() function.\
Codes are as follows:

```
// 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 {
int right;
bool flag;
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.
        flag= Iterator::hasNext();
       right  = Iterator::next();
	}

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

	// hasNext() and next() should behave the same as in the Iterator interface.
	// Override them if needed.
	int next() {
        int cur=right;
        flag = Iterator::hasNext();
        if(flag) 
        {
           right  = Iterator::next();
        }
	    return cur;
	}

	bool hasNext() const {
	    return flag;
	}
};
```

> 执行用时 :8 ms, 在所有 C++ 提交中击败了64.25%的用户\
> 内存消耗 :9.9 MB, 在所有 C++ 提交中击败了38.30%的用户


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