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描述
Design the CombinationIterator class:
- CombinationIterator(string characters, int combinationLength) Initializes the object with a string characters of sorted distinct lowercase English letters and a number combinationLength as arguments.
- next() Returns the next combination of length combinationLength in lexicographical order.
- hasNext() Returns true if and only if there exists a next combination.
Example 1:
Input
["CombinationIterator", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[["abc", 2], [], [], [], [], [], []]
Output
[null, "ab", true, "ac", true, "bc", false]
Explanation
CombinationIterator itr = new CombinationIterator("abc", 2);
itr.next(); // return "ab"
itr.hasNext(); // return True
itr.next(); // return "ac"
itr.hasNext(); // return True
itr.next(); // return "bc"
itr.hasNext(); // return False
Note:
1 <= combinationLength <= characters.length <= 15
All the characters of characters are unique.
At most 10^4 calls will be made to next and hasNext.
It's guaranteed that all calls of the function next are valid.
解析
根据题意，设计一个 CombinationIterator 类，里面需要包涵几个函数：
- CombinationIterator(string characters, int combineLength) 使用已经经过排序的不同小写英文字母的字符串 characters 和数字 combinationLength 作为参数初始化对象
- next() 按字典顺序返回长度为 combinationLength 的下一个组合
- hasNext() 当且仅当存在下一个组合时才返回 true
这道题的关键在于找出所有的组合，至于 next 和 hasNext 两个函数只需要做简单的逻辑判断即可，第一种方法使用了内置函数 itertools.combinations 来找出所有长度为 combinationLength 的组合，简单粗暴。
解答
class CombinationIterator(object):
def __init__(self, characters, combinationLength):
"""
:type characters: str
:type combinationLength: int
"""
self.characters = characters
self.L = [''.join(i) for i in itertools.combinations(characters, combinationLength)]
def next(self):
"""
:rtype: str
"""
return self.L.pop(0)
def hasNext(self):
"""
:rtype: bool
"""
if self.L:
return True
return False
运行结果
Runtime: 44 ms, faster than 81.82% of Python online submissions for Iterator for Combination.
Memory Usage: 16 MB, less than 72.73% of Python online submissions for Iterator for Combination.
解析
当然了使用内置函数显得没有水平，我们还可以自己写代码，自定义函数 permute 使用 DFS 找出所有的组合。
解答
class CombinationIterator(object):
def __init__(self, characters, combinationLength):
"""
:type characters: str
:type combinationLength: int
"""
self.characters = characters
self.n = combinationLength
self.N = len(characters)
self.L = []
self.permute('', 0)
def permute(self, s, start):
if len(s) == self.n:
self.L.append(s)
return
else:
for i in range(start, self.N):
self.permute(s + self.characters[i], i + 1)
def next(self):
"""
:rtype: str
"""
return self.L.pop(0)
def hasNext(self):
"""
:rtype: bool
"""
if self.L:
return True
return False
运行结果
Runtime: 52 ms, faster than 54.55% of Python online submissions for Iterator for Combination.
Memory Usage: 15.9 MB, less than 72.73% of Python online submissions for Iterator for Combination.
原题链接：leetcode.com/problems/it…
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