Peeking Iterator LeetCode Solution

Problem – Peeking Iterator LeetCode Solution

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 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 nexthasNext, and peek.

Follow up: How would you extend your design to be generic and work with all types, not just integer?

Peeking Iterator LeetCode Solution in Java

class PeekingIterator implements Iterator<Integer> {  
    private Integer next = null;
    private Iterator<Integer> iter;

    public PeekingIterator(Iterator<Integer> iterator) {
        // initialize any member here.
        iter = iterator;
        if (iter.hasNext())
            next = iter.next();
    }
    
    // Returns the next element in the iteration without advancing the iterator. 
    public Integer peek() {
        return next; 
    }

    // hasNext() and next() should behave the same as in the Iterator interface.
    // Override them if needed.
    @Override
    public Integer next() {
        Integer res = next;
        next = iter.hasNext() ? iter.next() : null;
        return res; 
    }

    @Override
    public boolean hasNext() {
        return next != null;
    }
}

Peeking Iterator LeetCode Solution in C++

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();
    }
};

Peeking Iterator LeetCode Solution in Python

class PeekingIterator(object):
    def __init__(self, iterator):
        self.iter = iterator
        self.temp = self.iter.next() if self.iter.hasNext() else None

    def peek(self):
        return self.temp

    def next(self):
        ret = self.temp
        self.temp = self.iter.next() if self.iter.hasNext() else None
        return ret

    def hasNext(self):
        return self.temp is not None
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