非常有趣的问题，但是这个“hasnext”的设计已经放进了leetcode: https://leetcode.com/problems/iterator-for-combination/

这是我的实现:

class CombinationIterator:

def __init__(self, characters: str, combinationLength: int):
    from itertools import combinations
    from collections import deque
    self.iter = combinations(characters, combinationLength)
    self.res = deque()


def next(self) -> str:
    if len(self.res) == 0:
        return ''.join(next(self.iter))
    else:
        return ''.join(self.res.pop())


def hasNext(self) -> bool:
    try:
        self.res.insert(0, next(self.iter))
        return True
    except:
        return len(self.res) > 0

非常有趣的问题，但是这个“hasnext”的设计已经放进了leetcode: https://leetcode.com/problems/iterator-for-combination/

这是我的实现:

class CombinationIterator:

def __init__(self, characters: str, combinationLength: int):
    from itertools import combinations
    from collections import deque
    self.iter = combinations(characters, combinationLength)
    self.res = deque()


def next(self) -> str:
    if len(self.res) == 0:
        return ''.join(next(self.iter))
    else:
        return ''.join(self.res.pop())


def hasNext(self) -> bool:
    try:
        self.res.insert(0, next(self.iter))
        return True
    except:
        return len(self.res) > 0

你可以使用itertools来tee迭代器。在teed迭代器上检查StopIteration。

为了读取所有迭代，基于处理“StopIteration”执行的解决方法非常简单:

    end_cursor = False
    while not end_cursor:
        try:
            print(cursor.next())
        except StopIteration:
            print('end loop')
            end_cursor = True
        except:
            print('other exceptions to manage')
            end_cursor = True

hasNext在某种程度上转换为StopIteration异常，例如:

>>> it = iter("hello")
>>> it.next()
'h'
>>> it.next()
'e'
>>> it.next()
'l'
>>> it.next()
'l'
>>> it.next()
'o'
>>> it.next()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
StopIteration

StopIteration文档:http://docs.python.org/library/exceptions.html#exceptions.StopIteration 一些关于python中的迭代器和生成器的文章:http://www.ibm.com/developerworks/library/l-pycon.html

在任意迭代器对象中尝试__length_hint__()方法:

iter(...).__length_hint__() > 0