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fizzbuzz

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力扣 412. Fizz Buzz
题目描述给你一个整数 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：
【fizzbuzz】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 <= 10^4解决方案方法一：模拟 + 字符串拼接题目要求返回数组 answer，对于每个 1≤i≤n，answer[i] 为 i 的转化。注意下标 i 从 1 开始。
根据题目描述，当 i 是 3 的倍数时 answer[i] 包含 “Fizz"，当 i 是 5 的倍数时 answer[i] 包含 “Buzz"，当 i 同时是 3 的倍数和 5 的倍数时 answer[i] 既包含 “Fizz" 也包含 “Fuzz"，且 “Fizz" 在 “Buzz" 前面。
基于上述规则，对于每个 1≤i≤n，可以使用字符串拼接的方式得到 answer[i]，具体操作如下：
如果 i 是 3 的倍数，则将 “Fizz" 拼接到 answer[i]；如果 i 是 5 的倍数，则将 “Buzz" 拼接到 answer[i]；如果 answer[i] 为空，则 i 既不是 3 的倍数也不是 5 的倍数，将 i 拼接到 answer[i] 。代码
Java

class Solution {public List<String> fizzBuzz(int n) {List<String> answer = new ArrayList<String>();for (int i = 1; i <= n; i++) {StringBuffer sb = new StringBuffer();if (i % 3 == 0) {sb.append("Fizz");}if (i % 5 == 0) {sb.append("Buzz");}if (sb.length() == 0) {sb.append(i);}answer.add(sb.toString());}return answer;}}

public class Solution {public IList<string> FizzBuzz(int n) {IList<string> answer = new List<string>();for (int i = 1; i <= n; i++) {StringBuilder sb = new StringBuilder();if (i % 3 == 0) {sb.Append("Fizz");}if (i % 5 == 0) {sb.Append("Buzz");}if (sb.Length == 0) {sb.Append(i);}answer.Add(sb.ToString());}return answer;}}

JavaScript

var fizzBuzz = function(n) {const answer = [];for (let i = 1; i <= n; i++) {const sb = [];if (i % 3 === 0) {sb.push("Fizz");}if (i % 5 === 0) {sb.push("Buzz");}if (sb.length === 0) {sb.push(i);}answer.push(sb.join(''));}return answer;};

C++

class Solution {public:vector<string> fizzBuzz(int n) {vector<string> answer;for (int i = 1; i <= n; i++) {string curr;if (i % 3 == 0) {curr　+= "Fizz";}if (i % 5 == 0) {curr += "Buzz";}if (curr.size() == 0) {curr += to_string(i);}answer.emplace_back(curr);}return answer;}};

Golang

func fizzBuzz(n int) (ans []string) {for i := 1; i <= n; i++ {sb := &strings.Builder{}if i%3 == 0 {sb.WriteString("Fizz")}if i%5 == 0 {sb.WriteString("Buzz")}if sb.Len() == 0 {sb.WriteString(strconv.Itoa(i))}ans = append(ans, sb.String())}return}

Python3

class Solution:def fizzBuzz(self, n: int) -> List[str]:ans = []for i in range(1, n + 1):s = ""if i % 3 == 0:s += "Fizz"if i % 5 == 0:s += "Buzz"if s == "":s = str(i)ans.append(s)return ans