Peeking Iterator Problem
Description
LeetCode Problem 284.
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:
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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.
Sample C++ Code
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/*
* Below is the interface for Iterator, which is already defined for you.
* **DO NOT** modify the interface for Iterator.
*
* class Iterator {
* struct Data;
* Data* data;
* public:
* Iterator(const vector<int>& nums);
* Iterator(const Iterator& iter);
*
* // Returns the next element in the iteration.
* int next();
*
* // Returns true if the iteration has more elements.
* bool hasNext() const;
* };
*/
class PeekingIterator : public Iterator
{
public:
PeekingIterator(const vector<int> &nums) : Iterator(nums)
{
}
int peek()
{
return Iterator(*this).next();
}
int next()
{
return Iterator::next();
}
bool hasNext() const
{
return Iterator::hasNext();
}
};