leetcode412：Fizz Buzz

carloscn commented 1 year ago

问题描述

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：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

来源：力扣（LeetCode）链接：https://leetcode.cn/problems/fizz-buzz

carloscn commented 1 year ago

问题分析：

根据条件进行判断就好了。

static int32_t fizz_buzz(size_t n)
{
    int32_t ret = 0;
    STRLIST_T *strlist = NULL;
    char buf[5] = {0};

    UTILS_CHECK_LEN(n);

    strlist = strlist_malloc();
    UTILS_CHECK_PTR(strlist);

    for (size_t i = 1; i < n + 1; i ++) {
        if ((i % 3 == 0) && (i % 5 != 0)) {
            strlist_add(strlist, "Fizz");
        } else if ((i % 5 == 0) && (i % 3 != 0)) {
            strlist_add(strlist, "Buzz");
        } else if ((i % 15) == 0) {
            strlist_add(strlist, "FizzBuzz");
        } else {
            sprintf(buf, "%zd", i);
            strlist_add(strlist, buf);
        }
    }
    strlist_infolog(strlist);

finish:
    strlist_free(strlist);
    return ret;
}

carloscn commented 1 year ago

code:

https://github.com/carloscn/structstudy/blob/master/c_programming/str/26_fizz-buzz_412.c

carloscn / structstudy