38 lines
1.7 KiB
Markdown
38 lines
1.7 KiB
Markdown
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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.
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Implement the `PeekingIterator` class:
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* `PeekingIterator(Iterator<int> nums)` Initializes the object with the given integer iterator `iterator`.
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* `int next()` Returns the next element in the array and moves the pointer to the next element.
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* `boolean hasNext()` Returns `true` if there are still elements in the array.
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* `int peek()` Returns the next element in the array **without** moving the pointer.
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**Note:** Each language may have a different implementation of the constructor and `Iterator`, but they all support the `int next()` and `boolean hasNext()` functions.
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**Example 1:**
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Input
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["PeekingIterator", "next", "peek", "next", "next", "hasNext"]
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[[[1, 2, 3]], [], [], [], [], []]
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Output
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[null, 1, 2, 2, 3, false]
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Explanation
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PeekingIterator peekingIterator = new PeekingIterator([1, 2, 3]); // [1,2,3]
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peekingIterator.next(); // return 1, the pointer moves to the next element [1,2,3].
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peekingIterator.peek(); // return 2, the pointer does not move [1,2,3].
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peekingIterator.next(); // return 2, the pointer moves to the next element [1,2,3]
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peekingIterator.next(); // return 3, the pointer moves to the next element [1,2,3]
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peekingIterator.hasNext(); // return False
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**Constraints:**
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* `1 <= nums.length <= 1000`
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* `1 <= nums[i] <= 1000`
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* All the calls to `next` and `peek` are valid.
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* At most `1000` calls will be made to `next`, `hasNext`, and `peek`.
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**Follow up:** How would you extend your design to be generic and work with all types, not just integer?
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https://leetcode.com/problems/peeking-iterator
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