为何无法用IF/ELSE比较STR与INT及FizzBuzz问题解法

问题需求 #

给定整数n，创建一个列表，遵循以下规则生成元素：

• 若数字同时被3和5整除，替换为"FizzBuzz"
• 若数字仅被3整除，替换为"Fizz"
• 若数字仅被5整除，替换为"Buzz"
• 其余数字保持原值，最终所有元素转为字符串形式

示例：

• n=3时返回["1","2","Fizz"]
• n=5时返回["1","2","Fizz","4","Buzz"]

最初困境与错误原因 #

在最初实现时，用IF/ELSE语句替换符合条件的数字后，后续判断能否被5整除时触发了TypeError: not all arguments converted during string formatting错误。
原因是：当把整数替换为字符串（比如把3换成"Fizz"）后，后续遍历到该元素时，尝试对字符串做取模运算（%）或者与整数比较，会因类型不匹配抛出错误——字符串无法和整数进行算术运算或直接比较。

低效解决方案 #

为了避开类型错误，先记录所有需要替换的元素，再循环替换对应位置的元素。虽然代码能运行，但多次遍历列表和元素查找导致效率较低：

def fizzbuzz(n):
    answer = []
    fizz_lst = []
    buzz_lst = []
    fizzbuzz_lst = []
    for i in range(n):
        answer.append(i+1)
        if answer[i] % 3 == 0 and answer[i] % 5 == 0:
            fizzbuzz_lst.append(answer[i])
        if answer[i] % 3 == 0:
            fizz_lst.append(answer[i])
        if answer[i] % 5 == 0:
            buzz_lst.append(answer[i])
        for i in range(len(fizzbuzz_lst)):
            if fizzbuzz_lst[i] in answer:
                ourindx = fizzbuzz_lst[i]
                answer[ourindx - 1] = "FizzBuzz"
        for i in range(len(buzz_lst)):
            if buzz_lst[i] in answer:
                ourindx = buzz_lst[i]
                answer[ourindx - 1] = "Buzz"
        for i in range(len(fizz_lst)):
            if fizz_lst[i] in answer:
                ourindx = fizz_lst[i]
                answer[ourindx - 1] = "Fizz"
    int_str = [str(x) for x in answer]
    return int_str

优化后的解决方案 #

调整思路：遍历列表时跳过已被替换为字符串的元素，仅对整数类型的元素进行IF/ELIF判断替换，从根源上避免类型不匹配的错误，同时减少不必要的操作，提升效率：

class Solution(object):
    def fizzBuzz(self, n):
        answer = []
        for i in range(n):
            answer.append(i + 1)
        for every_iter in answer:
            if type(every_iter) != int:
                continue
            elif every_iter % 3 == 0 and every_iter % 5 == 0:
                answer[every_iter-1] = "FizzBuzz"
            elif every_iter % 3 == 0:
                answer[every_iter-1] = "Fizz"
            elif every_iter % 5 == 0:
                answer[every_iter-1] = "Buzz"
        answer_str = [str(j) for j in answer]
        return answer_str

内容的提问来源于stack exchange，提问作者Deniz