# 3 Different ways to print Fibonacci series in Java

Given a number N, we need to find the Fibonacci Series up to the N term.

The Fibonacci series is a series of elements where, the previous two elements are added to get the next element, starting with 0 and 1.

Examples:

Input: N = 10
Output: 0 1 1 2 3 5 8 13 21 34
Here first term of Fibonacci is 0 and second is 1, so that 3rd term = first(o) + second(1) etc and so on.

Input: N = 15
Output: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377

Method 1 – Iterative: Initialize the first and second numbers to 0 and 1. Following this, we print the first and second numbers. Then we send the flow to the iterative while loop where we get the next number by adding the previous two number and simultaneously we swap the first number with the second and the second with the third.

Below is the implementation of the above approach:

## Java

 `// Java program for the above approach ` ` `  `class` `GFG { ` ` `  `    ``// Function to print N Fibonacci Number ` `    ``static` `void` `Fibonacci(``int` `N) ` `    ``{ ` `        ``int` `num1 = ``0``, num2 = ``1``; ` ` `  `        ``int` `counter = ``0``; ` ` `  `        ``// Iterate till counter is N ` `        ``while` `(counter < N) { ` ` `  `            ``// Print the number ` `            ``System.out.print(num1 + ``" "``); ` ` `  `            ``// Swap ` `            ``int` `num3 = num2 + num1; ` `            ``num1 = num2; ` `            ``num2 = num3; ` `            ``counter = counter + ``1``; ` `        ``} ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``// Given Number N ` `        ``int` `N = ``10``; ` ` `  `        ``// Function Call ` `        ``Fibonacci(N); ` `    ``} ` `} `

Output:

```0 1 1 2 3 5 8 13 21 34
```

Time Complexity: O(N)
Auxiliary Space: O(1)

Method 2 – Using Recursion: Since Fibonacci Number is the summation of the two previous numbers. We can use recursion as per the following condition:

1. Get the number whose Fibonacci series needs to be calculated.
2. Recursively iterate from value N to 1:
• Base case: If the value called recursively is less than 1, the return 1 the function.
• Recursive call: If the base case is not met, then recursively call for previous two value as:

recursive_function(N – 1) + recursive_function(N – 2);

• Return statement: At each recursive call(except the base case), return the recursive function for the previous two value as:

recursive_function(N – 1) + recursive_function(N – 2);

Below is the implementation of the above approach:

## Java

 `// Recursive implementation of ` `// Fibonacci Series ` ` `  `class` `GFG { ` ` `  `    ``// Function to print the fibonacci series ` `    ``static` `int` `fib(``int` `n) ` `    ``{ ` `        ``// Base Case ` `        ``if` `(n <= ``1``) ` `            ``return` `n; ` ` `  `        ``// Recursive call ` `        ``return` `fib(n - ``1``) ` `            ``+ fib(n - ``2``); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `    ``main(String args[]) ` `    ``{ ` `        ``// Given Number N ` `        ``int` `N = ``10``; ` ` `  `        ``// Print the first N numbers ` `        ``for` `(``int` `i = ``0``; i < N; i++) { ` ` `  `            ``System.out.print(fib(i) + ``" "``); ` `        ``} ` `    ``} ` `} `

Output:

```0 1 1 2 3 5 8 13 21 34
```

Time Complexity: O(2N)
Auxiliary Space: O(1)

Method 3 – Using Dynamic Programming: We can avoid the repeated work done in method 2 by storing the Fibonacci numbers calculated so far. Below are the steps:

1. Create an array arr[] of size N.
2. Initialize arr = 0, arr = 1.
3. Iterate over [2, N] and update the array arr[] as:

arr[i] = arr[i – 2] + arr[i – 1]

4. Print the value of arr[N].

Below is the implementation of the above approach:

## Java

 `// Dynamic Programming approach for ` `// Fibonacci Series ` ` `  `class` `fibonacci { ` ` `  `    ``// Function to find the fibonacci Series ` `    ``static` `int` `fib(``int` `n) ` `    ``{ ` ` `  `        ``// Declare an array to store ` `        ``// Fibonacci numbers. ` `        ``// 1 extra to handle case, n = 0 ` `        ``int` `f[] = ``new` `int``[n + ``2``]; ` ` `  `        ``int` `i; ` ` `  `        ``// 0th and 1st number of ` `        ``// the series are 0 and 1 ` `        ``f[``0``] = ``0``; ` `        ``f[``1``] = ``1``; ` ` `  `        ``for` `(i = ``2``; i <= n; i++) { ` ` `  `            ``// Add the previous 2 numbers ` `            ``// in the series and store it ` `            ``f[i] = f[i - ``1``] + f[i - ``2``]; ` `        ``} ` ` `  `        ``// Nth Fibonacci Number ` `        ``return` `f[n]; ` `    ``} ` ` `  `    ``public` `static` `void` `    ``main(String args[]) ` `    ``{ ` `        ``// Given Number N ` `        ``int` `N = ``10``; ` ` `  `        ``// Print first 10 term ` `        ``for` `(``int` `i = ``0``; i < N; i++) ` `            ``System.out.print(fib(i) + ``" "``); ` `    ``} ` `} `

Output:

```0 1 1 2 3 5 8 13 21 34
```

Time Complexity: O(N)
Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.