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
Solution
class PeekingIterator:
def __init__(self, iterator):
"""
Initialize your data structure here.
:type iterator: Iterator
"""
self.iterator = iterator
self.curr = self.iterator.next() if self.iterator.hasNext() else None
def peek(self):
"""
Returns the next element in the iteration without advancing the iterator.
:rtype: int
"""
return self.curr
def next(self):
"""
:rtype: int
"""
tmp = self.curr
self.curr = self.iterator.next() if self.iterator.hasNext() else None
return tmp
def hasNext(self):
"""
:rtype: bool
"""
return self.curr != None
if __name__ == '__main__':
p = PeekingIterator([1, 2 , 3])
p.next()
print(p.peek())
p.next()
p.next()
p.hasNext()