题目：原题链接（中等）
标签：数学、递归
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( 1 0 N ) | O ( 1 0 N ) | 164ms (8.93%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一（递归）：
class Solution:
def __init__(self):
self.reverse_lst = {
"0": "0",
"1": "1",
"6": "9",
"8": "8",
"9": "6"
}
self.ans = []
def findStrobogrammatic(self, n: int) -> List[str]:
self.count(n, "")
return self.ans
def count(self, n: int, now: str, first: bool = True):
if n == 0:
self.ans.append(now)
elif n <= len(now):
self.count(n - 1, now + self.reverse_lst[now[n - 1]], first=False)
elif n == 1 or n == len(now) + 1:
for num in ["0", "1", "8"]:
self.count(n - 1, now + num, first=False)
else:
if first:
lst = ["1", "6", "8", "9"]
else:
lst = ["0", "1", "6", "8", "9"]
for num in lst:
self.count(n - 1, now + num, first=False)
