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Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.35

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?