2000python
Posted on April 1, 2023
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
一道很简单的题，🥣。
题解:
impl Solution {
pub fn fizz_buzz(n: i32) -> Vec<String> {
let mut result:Vec<String> = Vec::new();
for x in 1..=n {
if x % 3 == 0 && x % 5 == 0 {
result.push("FizzBuzz".to_string());
}else if x % 3 == 0 {
result.push("Fizz".to_string());
}else if x % 5 == 0 {
result.push("Buzz".to_string());
}else {
result.push(x.to_string());
}
}
result
}
}
💖 💪 🙅 🚩
2000python
Posted on April 1, 2023
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