LEETCODE--Fizz Buzz

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本文介绍了一个简单的程序设计问题——FizzBuzz的实现方案。该程序输出从1到n的数字，但遇到3的倍数输出Fizz，5的倍数输出Buzz，同时为3和5的倍数则输出FizzBuzz。通过C++代码示例展示了如何使用条件判断和字符串流来实现这一功能。

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”
]

注意：
在进行多次转换的时候，必须调用stringstream的成员函数clear().

class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string> vec;
        string str;
        stringstream stream;
        for(int i = 1; i <= n; i++){
            if((i % 3 == 0) && (i % 5 == 0))
                str = "FizzBuzz";
            else if(i % 3 == 0)
                    str = "Fizz";
            else if(i % 5 == 0)
                    str = "Buzz";
            else{
                stream << i;
                stream >> str;
                stream.clear();//在进行多次转换前，必须清除stream

            }
            vec.push_back(str);
        }
        return vec;
    }
};