力扣4_412. Fizz Buzz

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"]

来源:力扣(LeetCode)

Java解法

class Solution {
    public List<String> fizzBuzz(int n) {
        List<String> lst = new ArrayList<String>();
        for (int i = 0; i < n; i++) {
            int num = i + 1;
            if (num % 3 == 0 && num % 5 == 0) lst.add("FizzBuzz");
            else if (num % 3 == 0) lst.add("Fizz");
            else if (num % 5 == 0) lst.add("Buzz");
            else lst.add(String.valueOf(num));
            //String.valueOf(num)将num转化成字符串
        }
        return lst;
    }
}

Python解法

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

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