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# Peeking Iterator LeetCode Solution

## Problem – Peeking Iterator LeetCode Solution

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.

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``````

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`.

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

## Peeking Iterator LeetCode Solution in Java

``````class PeekingIterator implements Iterator<Integer> {
private Integer next = null;
private Iterator<Integer> iter;

public PeekingIterator(Iterator<Integer> iterator) {
// initialize any member here.
iter = iterator;
if (iter.hasNext())
next = iter.next();
}

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

// hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
@Override
public Integer next() {
Integer res = next;
next = iter.hasNext() ? iter.next() : null;
return res;
}

@Override
public boolean hasNext() {
return next != null;
}
}
``````

## Peeking Iterator LeetCode Solution in C++

``````class PeekingIterator : public Iterator
{
public:
PeekingIterator(const vector<int> &nums) : Iterator(nums)
{
}

int peek()
{
return Iterator(*this).next();
}

int next()
{
return Iterator::next();
}

bool hasNext() const
{
return Iterator::hasNext();
}
};
``````

## Peeking Iterator LeetCode Solution in Python

``````class PeekingIterator(object):
def __init__(self, iterator):
self.iter = iterator
self.temp = self.iter.next() if self.iter.hasNext() else None

def peek(self):
return self.temp

def next(self):
ret = self.temp
self.temp = self.iter.next() if self.iter.hasNext() else None
return ret

def hasNext(self):
return self.temp is not None
``````
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