Problem:
Given an int n, return the string form of the number followed by "!". So the int 6 yields "6!". Except if the number is divisible by 3 use "Fizz" instead of the number, and if the number is divisible by 5 use "Buzz", and if divisible by both 3 and 5, use "FizzBuzz". Note: the % "mod" operator computes the remainder after division, so 23 % 10 yields 3. What will the remainder be when one number divides evenly into another? (See also: FizzBuzz Code and Introduction to Mod)
fizzString2(1) → "1!"
fizzString2(2) → "2!"
fizzString2(3) → "Fizz!"
Solution:
public String fizzString2(int n) {
boolean fizz = n % 3 == 0;
boolean buzz = n % 5 == 0;
if (fizz && buzz) return "FizzBuzz!";
if (fizz) return "Fizz!";
if (buzz) return "Buzz!";
return n + "!";
}
Bad Idea.
ReplyDeleteYour solution, is not suitable for change requirement.
What if you need to include divisible by 7 with a different word.
You need to change maximum lines.
Try a solution, where you need only one change, if there is one additional requirement.
public String fizzString2(int n) {
ReplyDeleteif(n%3==0 && n%5==0)
return "FizzBuzz!";
if(n%3==0)
return "Fizz!";
if(n%5==0)
return "Buzz!";
return n+"!";
}
private final String fizzString2(int fizzString) {
ReplyDeleteif(fizzString % 3 == 0){
if(fizzString % 5 == 0){
return "FizzBuzz!";
}
return "Fizz!";
}
if(fizzString % 5 == 0){
return "Buzz!";
}
return Integer.toString(fizzString) + "!";
}
if(n%5==0&&n%3==0)return "FizzBuzz!";
ReplyDeleteelse if(n%5==0&&n%3!=0) return "Buzz!";
else if(n%5!=0&& n%3==0) return "Fizz!";
else return n+"!";
String newStr = "" ;
ReplyDeleteif (n % 3 != 0 && n % 5 != 0) return n + "!" ;
if ( n % 3 == 0) newStr = newStr + "Fizz" ;
if (n % 5 == 0) newStr = newStr + "Buzz" ;
return newStr + "!" ;