284. Peeking Iterator
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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 iteratoriterator.int next()Returns the next element in the array and moves the pointer to the next element.boolean hasNext()Returnstrueif 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 <= 10001 <= nums[i] <= 1000- All the calls to
nextandpeekare valid. - At most
1000calls will be made tonext,hasNext, andpeek.
Follow up: How would you extend your design to be generic and work with all types, not just integer?
Hints
Hint 1
Think of "looking ahead". You want to cache the next element.
Hint 2
Is one variable sufficient? Why or why not?
Hint 3
Test your design with call order of
peek() before next() vs next() before peek().Hint 4
For a clean implementation, check out Google's guava library source code.