【leetcode系列】【py3】【简单】外观数列
题目:
原题链接: https://leetcode-cn.com/problems/count-and-say/
解题思路:
第一行已知为'1'
从第二行开始,递归遍历处理上一行字符串
直到第n行
代码实现:
class Solution:
def countAndSay(self, n: int) -> str:
result_str, floor = '1', 1
while floor < n:
num, tmp_str = 1, ""
for index in range(1, len(result_str)):
if result_str[index] == result_str[index - 1]:
num = num + 1
else:
tmp_str += str(num) + result_str[index - 1]
num = 1
tmp_str += str(num) + result_str[-1]
result_str = tmp_str
floor += 1
return result_str