题目：原题链接（中等）
标签：回溯算法、设计
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | 构造 = O ( 1 ) ; next = O ( 1 ); hasNext = O ( 1 ) | O ( 1 ) | 60ms (85.15%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
from scipy.special import comb
class CombinationIterator:
def __init__(self, characters: str, combinationLength: int):
self.characters = characters
self.length = combinationLength
self.empty = len(self.characters) - self.length
# 生成迭代器
self.iterator = self.dfs(0, "")
self.now = 0
self.total = comb(len(self.characters), self.length)
def next(self) -> str:
self.now += 1
return next(self.iterator)
def hasNext(self) -> bool:
return self.now < self.total
def dfs(self, i, now):
if len(now) == self.length:
yield now
else:
for res in self.dfs(i + 1, now + self.characters[i]):
yield res
if i - len(now) < self.empty:
for res in self.dfs(i + 1, now):
yield res
