leetcode之Peeking Iterator

本题是可以peek一下下一个数字，但是不改变next的状态。所以保存一个中间变量，当需要peek的时候就是用那个变量，而需要next的时候则中间变量+ 1，这样就能获得本身需要的功能了。代码如下：

# 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.item = iterator
self.items = []
while self.item.hasNext():
self.items.append(self.item.next())
print self.items
self.num = 0

def peek(self):
"""
Returns the next element in the iteration without advancing the iterator.
:rtype: int
"""
if self.num >= len(self.items):
pass
else:
return self.items[self.num]

def next(self):
"""
:rtype: int
"""
if self.num >= len(self.items):
pass
else:
self.num = self.num + 1
return self.items[self.num - 1]

def hasNext(self):
"""
:rtype: bool
"""
if self.num >= len(self.items):
return False
else:
return True