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
金銀銅鐵5 分钟前
用 Python 实现 Take-Away 游戏copyer_xyf1 小时前
Agent 流程编排copyer_xyf1 小时前
Agent RAGcopyer_xyf1 小时前
【RAG】向量数据库:milvuscopyer_xyf2 小时前
Agent 记忆管理JieE2129 小时前
LeetCode 56. 合并区间|超清晰 JS 图解思路,面试高频区间题星云穿梭17 小时前
用Python写一个带图形界面的学生管理系统——完整教程Jack2017 小时前
HarmonyOS开发中错误处理策略:网络异常统一处理金銀銅鐵17 小时前
用 Pygame 实现 15 puzzle小小杨树18 小时前
读懂色彩:拍照调色不再难