Medium

Quiz

#284 Peeking Iterator

APPROACH

Extend an existing iterator with a peek operation alongside hasNext and next.

Implement the PeekingIterator class:

- PeekingIterator(Iterator<int> nums) Wraps the given integer iterator iterator.

- int next() Returns the next element and advances the iterator.

- boolean hasNext() Returns true if more elements remain.

- int peek() Returns the next element without advancing the iterator.

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]

Example 2:

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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What is the optimal approach for this problem?