小黑跟认识了十一年的老同学们少又一次的聚齐，大年初一团跑南二环，初三龙年跑，跟尚香看电影吃火锅跟深沉吃烤肉的leetcode之旅:38. 外观数列

本文主要是介绍小黑跟认识了十一年的老同学们少又一次的聚齐，大年初一团跑南二环，初三龙年跑，跟尚香看电影吃火锅跟深沉吃烤肉的leetcode之旅:38. 外观数列，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

小黑代码

class Solution:# 缓存字典map_ = {1: '1'}def countAndSay(self, n: int) -> str:# 看是否在缓存中if n in self.map_:return self.map_[n]# 递归向下n_1_res = self.countAndSay(n-1)# 回溯pre_ = None# 记录前一个字符的个数count = 0# 结果数组res = []for c in n_1_res:# 如果前一个字符不为空，则判断该字符与前一个字符是否相等if pre_:# 相等的情况if pre_ == c:count += 1# 不相等的情况，记录结果，count置1，pre_设成当前字符else:res.extend([str(count), pre_])count = 1pre_ = c# 第一个字符的情况else:pre_ = ccount = 1# 记录末尾的字符序列res.extend([str(count), pre_])# 将数组拼接成字符串res = ''.join(res)self.map_[n] = resreturn res

在这里插入图片描述

1075. 项目员工 I

sql:

# Write your MySQL query statement below
SELECTproject_id, ROUND(AVG(experience_years), 2) AS average_years 
FROMProject LEFT JOIN Employee ON Project.employee_id = Employee.employee_id
GROUP BYproject_id

pandas:

import pandas as pddef project_employees_i(project: pd.DataFrame, employee: pd.DataFrame) -> pd.DataFrame:data = pd.merge(left=project, right=employee, on='employee_id')data = data.groupby('project_id')['experience_years'].mean().reset_index()data = data.rename(columns = {'experience_years': 'average_years'}).round(2)return data