Given an Iterator class interface with methods:
hasNext(), design and implement a PeekingIterator that support the
peek() operation – it essentially peek() at the element that will be returned by the next call to next().
Assume that the iterator is initialized to the beginning of the list: [1,2,3]. Call next() gets you 1, the first element in the list. Now you call peek() and it returns 2, the next element. Calling next() after that still return 2. You call next() the final time and it returns 3, the last element. Calling hasNext() after that should return false.
Follow up: How would you extend your design to be generic and work with all types, not just integer?
实现一个迭代器类，实现几个函数，注意要按照已有的 Iterator 类来展开。
# Below is the interface for Iterator, which is already defined for you. # # class Iterator(object): # def __init__(self, nums): # """ # Initializes an iterator object to the beginning of a list. # :type nums: List[int] # """ # # def hasNext(self): # """ # Returns true if the iteration has more elements. # :rtype: bool # """ # # def next(self): # """ # Returns the next element in the iteration. # :rtype: int # """ class PeekingIterator(object): def __init__(self, iterator): """ Initialize your data structure here. :type iterator: Iterator """ self.iterator = iterator self.top = None if self.iterator.hasNext(): self.top = self.iterator.next() def peek(self): """ Returns the next element in the iteration without advancing the iterator. :rtype: int """ return self.top def next(self): """ :rtype: int """ res = self.top if self.iterator.hasNext(): self.top = self.iterator.next() else: self.top = None return res def hasNext(self): """ :rtype: bool """ if self.top: return True else: return False # Your PeekingIterator object will be instantiated and called as such: # iter = PeekingIterator(Iterator(nums)) # while iter.hasNext(): # val = iter.peek() # Get the next element but not advance the iterator. # iter.next() # Should return the same value as [val]. # Runtime: 20 ms, faster than 89.90% of Python online submissions for Peeking Iterator. # Memory Usage: 12 MB, less than 6.30% of Python online submissions for Peeking Iterator.