Finding fibonacci number series in java

Fibonacci series : a series of numbers in which each number ( Fibonacci number ) is the sum of the two preceding numbers. The simplest is the series 1, 1, 2, 3, 5, 8, 13,21,34 ,55,89 etc.

Solution1 using Recursion : 

/**code to get nth fibonacci number **/

public static int fibonacciNumbers(int n){

  if(n <= 0 ||  n == 1){

   return 1;

  } else {

    return fibonacciNumbers(n-1) + fibonacciNumbers(n-2);

   }

 }

public static void main(String[] args){

int numberOfFibonacci  = 15;

  long start = System.currentTimeMillis();

  java.util.List<Integer> fibnocList = new ArrayList<>();

  for (int i = 0; i <= numberOfFibonacci; i++) {

   fibnocList.add(fibonacciNumbers(i));

  }

  System.out.println(fibnocList);

  long end = System.currentTimeMillis();

  System.out.println("total time taken=" + (end -start)  +  " ms");

 }
Output : 

[1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987]




total time taken=1 ms

Note : Till 40th element its able to print fibonacci numbers after that time of execution

increase exponentially.

Time complexity is O(n) & Space complexity O(n).

Solution2 using  without Recursion : 

       
/**printing the series of fibonacci number upto length n  **/


public static void fibonacciSeries(int n){

  int[] fibArray = new int[n];

  if(n < 0 ){

   System.out.println("invalid value");

  } else {

     for (int i = 0; i < n; i++) {

      if(i== 0 || i==1){

       fibArray[i] =1;

      } else {

       fibArray[i] =  fibArray[i-1] + fibArray[i-2];

      }

   }

   }

  System.out.println("fibonacci number upto lenth n=" + n + " and series=" + Arrays.toString(fibArray));

 }

public static void main(String[] args){

  long start = System.currentTimeMillis();

  fibonacciSeries(20);

  long end = System.currentTimeMillis();

  System.out.println("total time taken=" + (end -start)  +  " ms");

 }

Note : Time complexity O(n) and space complexity is in place as using the arrays.

Output: 



fibonacci number upto lenth n=20 and series=[1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 

233, 377, 610, 987, 1597, 2584, 4181, 6765]

total time taken=0 ms
      

In both the solutions number 2 solutions is best possible solution with respect to time and space complexity.

Enjoy the code and enjoy the journey of Java!!!!