isPrime Problem

Author: JJ27

src/com/codefortomorrow/advanced/chapter13/practice/PrimePractice.java

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+package com.codefortomorrow.advanced.chapter13.practice;
+/*
+Write a method called isPrime which returns
+true if the given integer is prime and false otherwise.
+
+This is similar to the chapter 11 problem, but this time write
+your method using recursion.
+
+In your main method, include a scanner so the user can check
+as many numbers as they want until they enter -1.
+*/
+public class PrimePractice{
+
+}

src/com/codefortomorrow/advanced/chapter13/solutions/PrimePractice.class

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+package com.codefortomorrow.advanced.chapter13.practice;
+/*
+Write a method called isPrime which returns
+true if the given integer is prime and false otherwise.
+
+This is similar to the chapter 11 problem, but this time write
+your method using recursion.
+
+In your main method, include a scanner so the user can check
+as many numbers as they want until they enter -1.
+
+Note: There are more complex solutions, but this is the fastest one
+within the scope of this chapter.
+*/
+import java.util.Scanner;
+import java.lang.Math;
+public class PrimePractice{
+  public static void main(String[] args){
+    int s = 0;
+    while(s != -1){
+      Scanner reader = new Scanner(System.in);
+      System.out.print("Enter an integer to check: ");
+      s = reader.nextInt();
+      if(s != -1){
+        if(isPrime(s,2))
+          System.out.println("That is a prime!");
+        else
+          System.out.println("Not a prime!");
+      }
+    }
+  }
+  //n is the number to check, z is the current number being divided
+  public static boolean isPrime(int n, int z){
+    //Check base cases
+    if(n <= 2)
+      return (n == 2) ? true : false;
+      //Ternary operator used there
+    if(n % z == 0)
+      return false;
+    //If z gets high enough that z > sqrt(n), then n is prime, because factors just repeat after
+    if(Math.pow(z,2) > n)
+      return true;
+
+    //If none of the above work
+    return isPrime(n, z + 1);
+  }
+}