英文原文
Given an integer n, return a string array answer (1-indexed) where:
answer[i] == "FizzBuzz"ifiis divisible by3and5.answer[i] == "Fizz"ifiis divisible by3.answer[i] == "Buzz"ifiis divisible by5.answer[i] == iif non of the above conditions are true.
Example 1:
Input: n = 3 Output: ["1","2","Fizz"]
Example 2:
Input: n = 5 Output: ["1","2","Fizz","4","Buzz"]
Example 3:
Input: n = 15 Output: ["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]
Constraints:
1 <= n <= 104
中文题目
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
通过代码
高赞题解
模拟
根据题意进行模拟。
代码：
[]class Solution { public List<String> fizzBuzz(int n) { List<String> ans = new ArrayList<>(); for (int i = 1; i <= n; i++) { String cur = ""; if (i % 3 == 0) cur += "Fizz"; if (i % 5 == 0) cur += "Buzz"; if (cur.length() == 0) cur = i + ""; ans.add(cur); } return ans; } }
- 时间复杂度：$O(n)$
- 空间复杂度：$O(n)$
其他「模拟」相关内容
题目简单？考虑加餐一道 01 背包变形题。
或是考虑加练如下「模拟」题目 🍭🍭
注：以上目录整理来自 wiki，任何形式的转载引用请保留出处。
最后
如果有帮助到你，请给题解点个赞和收藏，让更多的人看到 ~ (“▔□▔)/
也欢迎你 关注我（公主号后台回复「送书」即可参与长期看题解学算法送实体书活动）或 加入「组队打卡」小群 ，提供写「证明」&「思路」的高质量题解。
所有题解已经加入 刷题指南，欢迎 star 哦 ~
统计信息
| 通过次数 | 提交次数 | AC比率 |
|---|---|---|
| 105793 | 149152 | 70.9% |
提交历史
| 提交时间 | 提交结果 | 执行时间 | 内存消耗 | 语言 |
|---|
