《LeetCode之每日一题》:177.Fizz Buzz

Posted 2021-11-02 是七喜呀！

Fizz Buzz

• 有关题目
• • 题解

题目链接： Fizz Buzz

有关题目

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 <= 10^4

题解

法一：模拟
代码一：
参考官方题解

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;
    }
};

代码二：

class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string> answer;
        for (int i = 1; i <= n; i++)
        {
            string curr;
            if (i % 15 == 0)
                curr += "FizzBuzz";
            else if (i % 3 == 0)
                curr += "Fizz";
            else if (i % 5 == 0)
                curr += "Buzz";
            else 
                curr += to_string(i);

            answer.emplace_back(curr);
        }
    
        return answer;
    }
};

代码三：
参考官方题解评论区下Kuroko

class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string> answer(n);
        int i;
        for (i = 1; i <= n; i++) answer[i - 1] = to_string(i);
        for (i = 3; i <= n; i += 3) answer[i - 1] = "Fizz";
        for (i = 5; i <= n; i += 5) answer[i - 1] = "Buzz";
        for (i = 15; i <= n; i += 15) answer[i - 1] = "FizzBuzz";

        return answer;
    }
};