What is the main pattern in Fizz Buzz that needs attention?

The key pattern is Math plus String: using modulo to detect multiples of 3, 5, or both, and converting results into strings accordingly.

Can I optimize Fizz Buzz beyond O(n)?

No, each number from 1 to n must be processed at least once to apply the replacement rules, so O(n) is the expected time complexity.

Why is the order of divisibility checks important?

Checking 3 and 5 separately before the combined check can overwrite 'FizzBuzz', causing incorrect results. Combined check should be prioritized or handled with concatenation.

How does GhostInterview help avoid off-by-one mistakes?

It ensures all loops and array accesses are correctly 1-indexed and alerts users when standard zero-based loops might cause errors.

Are there space optimizations for Fizz Buzz?

The main space use is the output array of size n; preallocation avoids dynamic resizing, and streaming approaches can reduce peak memory for very large n.

#412

Easy

auto_awesome数学·string

LeetCode 题解工作台

Fizz Buzz

给你一个整数 n ，返回一个字符串数组 answer （下标从 1 开始），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…

数学字符串模拟

题目描述

给你一个整数 n ，返回一个字符串数组 answer（下标从 1 开始），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

lightbulb

解题思路

方法一：模拟

我们遍历从 1 到 n 的每个整数，对于每个整数，我们检查它是否是 3 和 5 的倍数，或者只是 3 的倍数，或者只是 5 的倍数。根据检查的结果，我们将相应的字符串添加到答案数组中。

时间复杂度 $O(n)$ ，其中 $n$ 是题目给定的整数。忽略答案数组的空间消耗，空间复杂度 $O(1)$ 。

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class Solution:
    def fizzBuzz(self, n: int) -> List[str]:
        ans = []
        for i in range(1, n + 1):
            if i % 15 == 0:
                ans.append('FizzBuzz')
            elif i % 3 == 0:
                ans.append('Fizz')
            elif i % 5 == 0:
                ans.append('Buzz')
            else:
                ans.append(str(i))
        return ans

speed

复杂度分析

指标	值
时间	complexity is O(n) since each number is checked once for divisibility and string concatenation is constant time. Space complexity is O(n) for the output array, with no extra data structures required beyond simple string variables.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Checks for modulo operations versus combined conditional logic.
question_mark
Watches for correct 1-indexed output versus off-by-one errors.
question_mark
Notes clarity in string concatenation instead of multiple nested conditions.

warning

常见陷阱

外企场景

error
Using zero-based indexing incorrectly, leading to off-by-one results.
error
Checking divisibility in the wrong order, causing 'FizzBuzz' to be replaced by 'Fizz' or 'Buzz'.
error
Appending numbers instead of converting to strings for non-multiples.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Return the result as a comma-separated string instead of an array.
arrow_right_alt
Modify divisibility rules to use different multiples like 2 and 7 for custom patterns.
arrow_right_alt
Handle large n with streaming output to reduce memory usage.

help

常见问题

外企场景

继续练习

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