412. Fizz Buzz

最新推荐文章于 2024-11-23 18:16:46 发布

t_jianyan

最新推荐文章于 2024-11-23 18:16:46 发布

阅读量198

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode代码笔记

本文链接：https://blog.youkuaiyun.com/t_jianyan/article/details/53063632

leetcode代码笔记专栏收录该内容

82 篇文章

订阅专栏

本文介绍了一个简单的程序设计问题——FizzBuzz问题的实现方法。通过三种不同的C++代码示例展示了如何输出1到n的数字，并将3的倍数替换为Fizz，5的倍数替换为Buzz，同时为15的倍数输出FizzBuzz。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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”.

n = 15,

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

class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string> ans;
        for(int i = 1; i <= n; i++)
        {
            if(i%3 == 0)
            {
                if(i%5 == 0)
                ans.push_back("FizzBuzz");  //双引号 不是单引号
                else 
                ans.push_back("Fizz");
            }
            else if(i%5 == 0)
            ans.push_back("Buzz");
            else
            ans.push_back(to_string(i));    //  to_string 的用法
        }
        return ans;
    }
};

更好的

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

另一种用到res[i-1] 时之前的定义不能写成 vector<string> res; 要写成下面的样子。

class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string> res(n);
        for(int i=1; i<=n; ++i)
        {
            if(i%15==0)
                res[i-1] = "FizzBuzz";
            else if(i%5==0)
                res[i-1] = "Buzz";
            else if(i%3==0)
                res[i-1] = "Fizz";
            else
                res[i-1] = to_string(i);
        }
        return res;
    }
};