外观数列
外观数列
https://leetcode-cn.com/problems/count-and-say/
这题有意思
可以打表,不过打表的过程也相当于做出来了
class Solution:
def countAndSay(self,n: int) -> str:
if n == 1:
return '1'
s = self.countAndSay(n - 1)
n,res = 0,''
for ii,ss in enumerate(s):
if ss != s[n]:
res += str(ii-n) + s[n]
n = ii
res += str(len(s) - n) + s[-1]
return res
print(Solution().countAndSay(3))思路:
递归,将上一层计算出来的东西作为迭代对象。