Given an integer n, return a string array answer (1-indexed) where:
answer[i] == "FizzBuzz"ifiis divisible by3and5.answer[i] == "Fizz"ifiis divisible by3.answer[i] == "Buzz"ifiis divisible by5.answer[i] == i(as a string) if none of the above conditions are true.
Example 1:
Input: n = 3 Output: ["1","2","Fizz"]
Example 2:
Input: n = 5 Output: ["1","2","Fizz","4","Buzz"]
Example 3:
Input: n = 15 Output: ["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]
Constraints:
1 <= n <= 104
var fizzBuzz = function(n) {
const output = []
for(let i = 1; i <= n; i++) {
let str = i %% 3 === 0 ? 'Fizz' : ''
str += i %% 5 === 0 ? 'Buzz' : ''
output.push(str || String(i))
}
return output
};
這題其實太簡單了一點,非要說技巧的話就是,用分別把字串除 3 和 5,加入 Fizz 和 Bizz,可以少做兩個同時可以除盡的情形減少 loop 內的複雜度。
