Algorithm
Problem Name: 284. Peeking Iterator
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 iteratoriterator.int next()Returns the next element in the array and moves the pointer to the next element.boolean hasNext()Returnstrueif 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 <= 10001 <= nums[i] <= 1000- All the calls to
nextandpeekare valid. - At most
1000calls will be made tonext,hasNext, andpeek.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Iterator {
struct Data;
Data* data;
public:
Iterator(const vector<int>& nums);
Iterator(const Iterator& iter);
virtual ~Iterator();
// Returns the next element in the iteration.
int next();
// Returns true if the iteration has more elements.
bool hasNext() const;
};
class PeekingIterator : public Iterator {
private:
deque < int>q;
public:
PeekingIterator(const vector<int>& nums) : Iterator(nums) {}
// Returns the next element in the iteration without advancing the iterator.
int peek() {
if(q.empty()) q.push_front(Iterator::next());
return q.front();
}
// hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
int next() {
if(!q.empty()){
int t = q.front();
q.pop_front();
return t;
}
return Iterator::next();;
}
bool hasNext() const {
return Iterator::hasNext() || !q.empty();
}
};
Input
["PeekingIterator", "next", "peek", "next", "next", "hasNext"]
[[[1, 2, 3]], [], [], [], [], []]
Output
[null, 1, 2, 2, 3, false]
#2 Code Example with Java Programming
Code -
Java Programming
class PeekingIterator implements Iterator {
private Iterator iterator;
private Integer topElement;
public PeekingIterator(Iterator < Integer> iterator) {
// initialize any member here.
this.iterator = iterator;
populateTopElement();
}
// Returns the next element in the iteration without advancing the iterator.
public Integer peek() {
return this.topElement;
}
// hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
@Override
public Integer next() {
Integer nextValue = this.topElement;
populateTopElement();
return nextValue;
}
@Override
public boolean hasNext() {
return this.topElement != null;
}
private void populateTopElement() {
this.topElement = this.iterator.hasNext() ? this.iterator.next() : null;
}
}
Input
["PeekingIterator", "next", "peek", "next", "next", "hasNext"]
[[[1, 2, 3]], [], [], [], [], []]
Output
[null, 1, 2, 2, 3, false]
#3 Code Example with Python Programming
Code -
Python Programming
class PeekingIterator:
def __init__(self, iterator):
self.it = iterator
self.bool = self.it.hasNext()
self.num = self.it.next() if self.bool else None
def peek(self):
return self.num
def next(self):
n = self.num
self.bool = self.it.hasNext()
if self.bool:
self.num = self.it.next()
return n
def hasNext(self):
return self.bool
Input
["PeekingIterator", "next", "peek", "next", "next", "hasNext"]
[[[1, 2, 3]], [], [], [], [], []]
Output
[null, 1, 2, 2, 3, false]