FizzBuzz

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/*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”.*/
public class FizzBuzz {
public List fizzBuzz(int n) {
long startTime=System.currentTimeMillis();
List l=new ArrayList();

    for(int i=1;i<=n;i++){        if(i%3==0&&i%5!=0){            l.add("Fizz");        }else if(i%3!=0&&i%5==0){            l.add("Buzz");        }else if(i%3==0&&i%5==0){            l.add("FizzBuzz");        }else{            l.add(Integer.toString(i));        }    }    long endTime=System.currentTimeMillis(); //获取结束时间    System.out.println("程序运行时间： "+(endTime-startTime)+"ms");    return l;}//    解决方案2 如果取模为0则归零继续循环 速度较快 不用百分号%public List<String> fizzBuzz2(int n) {    long startTime=System.currentTimeMillis();    List<String> ret = new ArrayList<String>(n);    for(int i=1,fizz=0,buzz=0;i<=n ;i++){        fizz++;        buzz++;        if(fizz==3 && buzz==5){            ret.add("FizzBuzz");            fizz=0;            buzz=0;        }else if(fizz==3){            ret.add("Fizz");            fizz=0;        }else if(buzz==5){            ret.add("Buzz");            buzz=0;        }else{            ret.add(String.valueOf(i));        }    }    long endTime=System.currentTimeMillis(); //获取结束时间    System.out.println("程序运行时间： "+(endTime-startTime)+"ms");    return ret;}

// 速度慢
public List fizzBuzz3(int n) {
long startTime=System.currentTimeMillis();
List list = new ArrayList();
for (int i = 1; i <= n; i++) {
String item = “”;
if (i % 3 == 0) {
item += “Fizz”;
}
if (i % 5 == 0) {
item += “Buzz”;
}
if (item.equals(“”)) {
item = “” + i;
}
list.add(item);
}
long endTime=System.currentTimeMillis(); //获取结束时间

    System.out.println("程序运行时间： "+(endTime-startTime)+"ms");    return list;}public static void main(String[] args) {    FizzBuzz fb=new FizzBuzz();    fb.fizzBuzz(15000000);    fb.fizzBuzz2(15000000);    fb.fizzBuzz3(15000000);}

}

“`/*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”.*/
public class FizzBuzz {
public List fizzBuzz(int n) {
long startTime=System.currentTimeMillis();
List l=new ArrayList();

    for(int i=1;i<=n;i++){        if(i%3==0&&i%5!=0){            l.add("Fizz");        }else if(i%3!=0&&i%5==0){            l.add("Buzz");        }else if(i%3==0&&i%5==0){            l.add("FizzBuzz");        }else{            l.add(Integer.toString(i));        }    }    long endTime=System.currentTimeMillis(); //获取结束时间    System.out.println("程序运行时间： "+(endTime-startTime)+"ms");    return l;}//    解决方案2 如果取模为0则归零继续循环 速度较快 不用百分号%public List<String> fizzBuzz2(int n) {    long startTime=System.currentTimeMillis();    List<String> ret = new ArrayList<String>(n);    for(int i=1,fizz=0,buzz=0;i<=n ;i++){        fizz++;        buzz++;        if(fizz==3 && buzz==5){            ret.add("FizzBuzz");            fizz=0;            buzz=0;        }else if(fizz==3){            ret.add("Fizz");            fizz=0;        }else if(buzz==5){            ret.add("Buzz");            buzz=0;        }else{            ret.add(String.valueOf(i));        }    }    long endTime=System.currentTimeMillis(); //获取结束时间    System.out.println("程序运行时间： "+(endTime-startTime)+"ms");    return ret;}

    System.out.println("程序运行时间： "+(endTime-startTime)+"ms");    return list;}public static void main(String[] args) {    FizzBuzz fb=new FizzBuzz();    fb.fizzBuzz(15000000);    fb.fizzBuzz2(15000000);    fb.fizzBuzz3(15000000);}

}
很简单的取模循环，自己写了第一种，然后去discuss看了下别人的方案，发现了第二种解决方案不用百分号。速度上12相差不大，第三种最慢

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