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Java-based LeetCode algorithm problem solutions, regularly updated
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 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?