算法 简单 | 9. Fizz Buzz 问题

题目描述

给你一个整数n. 从 1 到 n 按照下面的规则打印每个数：

1. 如果这个数被3整除，打印fizz.
2. 如果这个数被5整除，打印buzz.
3. 如果这个数能同时被3和5整除，打印fizz buzz.
4. 如果这个数既不能被 3 整除也不能被 5 整除，打印数字本身。

样例

比如 n = 15, 返回一个字符串数组：
[
  "1", "2", "fizz",
  "4", "buzz", "fizz",
  "7", "8", "fizz",
  "buzz", "11", "fizz",
  "13", "14", "fizz buzz"
]

java题解

class Solution {
    /**
     * param n: As description.
     * return: A list of strings.
     */
    public ArrayList<String> fizzBuzz(int n) {
        ArrayList<String> results = new ArrayList<String>();
        for (int i = 1; i <= n; i++) {
            //能同时被3跟5整除的数，肯定也能被3整除，也能被5整除，所以要先判断能否被15整除
            if (i % 15 == 0) {
                results.add("fizz buzz");
            } else if (i % 5 == 0) {
                results.add("buzz");
            } else if (i % 3 == 0) {
                results.add("fizz");
            } else {
                results.add(String.valueOf(i));
            }
        }
        return results;
    }
}

C++题解

class Solution {
public:
    /*
     * param n: As description.
     * return: A list of strings.
     */
    vector<string> fizzBuzz(int n) {
        vector<string> results;
        for (int i = 1; i <= n; i++) {
            //能同时被3跟5整除的数，肯定也能被3整除，也能被5整除，所以要先判断能否被15整除
            if (i % 15 == 0) {
                results.push_back("fizz buzz");
            } else if (i % 5 == 0) {
                results.push_back("buzz");
            } else if (i % 3 == 0) {
                results.push_back("fizz");
            } else {
                results.push_back(to_string(i));
            }
        }
        return results;
    }
};

python题解

class Solution:
    """
    @param n: An integer as description
    @return: A list of strings.
    For example, if n = 7, your code should return
        ["1", "2", "fizz", "4", "buzz", "fizz", "7"]
    """
    def fizzBuzz(self, n):
        results = []
        for i in range(1, n+1):
            # 能同时被3跟5整除的数，肯定也能被3整除，也能被5整除，所以要先判断能否被15整除
            if i % 15 == 0:
                results.append("fizz buzz")
            elif i % 5 == 0:
                results.append("buzz")
            elif i % 3 == 0:
                results.append("fizz")
            else:
                results.append(str(i))
        return results