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
多加点辣也没关系21 小时前
数据结构与算法|第六章:队列2401_8330336221 小时前
golang如何实现MQTT主题通配符路由_golang MQTT主题通配符路由实现策略AI精钢21 小时前
修复 AI Gateway 图片 MIME 类型错误:用魔数检测替代扩展名猜测m0_5967490921 小时前
Golang怎么实现方法集与接口的匹配_Golang如何理解值类型和指针类型实现接口的区别【详解】_深海凉_21 小时前
LeetCode热题100-分割回文串我是无敌小恐龙1 天前
Java基础入门Day10 | Object类、包装类、大数/日期类、冒泡排序与Arrays工具类 超详细总结隔壁小红馆1 天前
隐藏odoo特有ADI_OP1 天前
用SigmaStudio+软件来开发ADSP-21565lifewange1 天前
pytest 找不到文件?直接在 pytest.ini 配置根目录 + 路径(最简单方案)