LeetCode: 412 Fizz Buzz(easy)

玲珑子_a

于 2019-04-17 14:24:05 发布

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分类专栏： LeetCode学习笔记

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本文探讨了经典的编程挑战——FizzBuzz问题，并提供了两种不同的C++解决方案。一种是通过条件判断来决定输出，另一种则是逐步构建字符串。这两种方法都有效地解决了从1到n的数字输出问题，对于学习条件语句和字符串操作的初学者来说，是很好的实践案例。

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题目：

Write a program that outputs the string representation of numbers from 1 to n.

But for multiples of three it should output “Fizz” instead of the number and for the multiples of five output “Buzz”. For numbers which are multiples of both three and five output “FizzBuzz”.

Example:

n = 15,

Return:
[
    "1",
    "2",
    "Fizz",
    "4",
    "Buzz",
    "Fizz",
    "7",
    "8",
    "Fizz",
    "Buzz",
    "11",
    "Fizz",
    "13",
    "14",
    "FizzBuzz"
]

代码：

自己的：

class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string> result;
        for(int i = 1; i < n+1; i++){
            if ((i%3 == 0)&&(i%5 == 0))
                result.push_back("FizzBuzz");
            else if ((i%3 != 0)&&(i%5 == 0))
                result.push_back("Buzz");
            else if ((i%3 == 0)&&(i%5 != 0))
                result.push_back("Fizz");
            else
                result.push_back(to_string(i));
        }
        return result;
    }
};

别人的：

class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string> res;
        for(int i=1; i<=n; i++){
            string s;
            if(i%3==0) s = "Fizz";
            if(i%5==0) s += "Buzz";
            if(s.empty())   s = to_string(i);
            res.push_back(s);
        }
        return res;
    }
};