难度:简单
给你一个整数
n,找出从1到n各个整数的 Fizz Buzz 表示,并用字符串数组answer(下标从 1 开始)返回结果,其中:
answer[i] == "FizzBuzz"如果i同时是3和5的倍数。answer[i] == "Fizz"如果i是3的倍数。answer[i] == "Buzz"如果i是5的倍数。answer[i] == i(以字符串形式)如果上述条件全不满足。示例 1:
输入:n = 3 输出:["1","2","Fizz"]示例 2:
输入:n = 5 输出:["1","2","Fizz","4","Buzz"]示例 3:
输入:n = 15 输出:["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]提示:
1 <= n <= 104
题解:
pythonclass Solution(object): def fizzBuzz(self, n): res = [] for i in range(1,n+1): # print(i) if i % 3== 0 and i %5 == 0: res.append('FizzBuzz') elif i%3 == 0: res.append('Fizz') elif i%5 == 0: res.append('Buzz') else: res.append(str(i)) return res
leetcode:412. Fizz Buzz(python3解法)
心软且酷丶2024-01-10 21:31
相关推荐
mCell6 小时前
Lua 编程入门:从基础语法到元表安_7 小时前
如何构建和使用向量索引?HNSW 和 IVF 有什么区别?counting money8 小时前
Java IO流详解:从InputStream到文件操作实战wuyk5558 小时前
98.C语言易混难点:字符数组与字符串指针的底层差异坚持学习前端日记8 小时前
Python SQLAlchemy ORM 从0到1精通实战手册(基础到复杂高阶)峥无8 小时前
从0到1手撕红黑树:封装实现 my_map 与 my_set(SGI-STL 源码级深度解析)北斗落凡尘9 小时前
LangGraph 入门实战(11)--输出模式不会代码的小猴9 小时前
3. 控件学习1雪碧聊技术9 小时前
安装Python(保姆级教程)