284. Peeking Iterator
Description
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()
Returnstrue
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
andpeek
are valid. - At most
1000
calls 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?
Solutions
Solution 1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 |
|