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Fizz Buzz(C++)

池永长
2023-12-01

Fizz Buzz

难度简单224收藏分享切换为英文接收动态反馈

给你一个整数 n ,找出从 1n 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 35 的倍数。
  • answer[i] == "Fizz" 如果 i3 的倍数。
  • answer[i] == "Buzz" 如果 i5 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

我的代码:

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

对应我的掘金文章:https://juejin.cn/post/7147265285458755597

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