# 284. Peeking Iterator

#### Medium

***

Design an iterator that supports the `peek` operation on an existing iterator in addition to the `hasNext` and the `next` operations.

Implement the `PeekingIterator` class:

* `PeekingIterator(Iterator<int> nums)` Initializes the object with the given integer iterator `iterator`.
* `int next()` Returns the next element in the array and moves the pointer to the next element.
* `boolean hasNext()` Returns `true` if there are still elements in the array.
* `int peek()` Returns the next element in the array **without** moving the pointer.

**Note:** Each language may have a different implementation of the constructor and `Iterator`, but they all support the `int next()` and `boolean hasNext()` functions.

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**Example 1:**

```
Input
["PeekingIterator", "next", "peek", "next", "next", "hasNext"]
[[[1, 2, 3]], [], [], [], [], []]
Output
[null, 1, 2, 2, 3, false]

Explanation
PeekingIterator peekingIterator = new PeekingIterator([1, 2, 3]); // [1,2,3]
peekingIterator.next();    // return 1, the pointer moves to the next element [1,2,3].
peekingIterator.peek();    // return 2, the pointer does not move [1,2,3].
peekingIterator.next();    // return 2, the pointer moves to the next element [1,2,3]
peekingIterator.next();    // return 3, the pointer moves to the next element [1,2,3]
peekingIterator.hasNext(); // return False
```

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**Constraints:**

* `1 <= nums.length <= 1000`
* `1 <= nums[i] <= 1000`
* All the calls to `next` and `peek` are valid.
* At most `1000` calls will be made to `next`, `hasNext`, and `peek`.

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

#### Solution : Handles None values as well

```python
class PeekingIterator:
    def __init__(self, iterator):
        self._iterator = iterator
        self._current = None
        self._hasNext = True
        self.next()

    def peek(self):
        return self._current

    def next(self):
        current = self._current
        if self._iterator.hasNext():
            self._current = self._iterator.next()
        else:
            self._hasNext = False
            
        return current

    def hasNext(self):
        return self._hasNext
```

```python
# Below is the interface for Iterator, which is already defined for you.
#
# class Iterator:
#     def __init__(self, nums):
#         """
#         Initializes an iterator object to the beginning of a list.
#         :type nums: List[int]
#         """
#
#     def hasNext(self):
#         """
#         Returns true if the iteration has more elements.
#         :rtype: bool
#         """
#
#     def next(self):
#         """
#         Returns the next element in the iteration.
#         :rtype: int
#         """

class PeekingIterator:
    def __init__(self, iterator):
        """
        Initialize your data structure here.
        :type iterator: Iterator
        """
        self.iterator = iterator
        self.temp = self.iterator.next() if self.iterator.hasNext() else None
        

    def peek(self):
        """
        Returns the next element in the iteration without advancing the iterator.
        :rtype: int
        """
        return self.temp
        

    def next(self):
        """
        :rtype: int
        """
        value = self.temp
        self.temp = self.iterator.next() if self.iterator.hasNext() else None
        return value
        

    def hasNext(self):
        """
        :rtype: bool
        """
        return self.temp is not None
        

# Your PeekingIterator object will be instantiated and called as such:
# iter = PeekingIterator(Iterator(nums))
# while iter.hasNext():
#     val = iter.peek()   # Get the next element but not advance the iterator.
#     iter.next()         # Should return the same value as [val].
```