Fibonacci Sequence in Java taking too long?

Question

I\'m trying to find the sum of the Fibonacci sequence in Java, but the run time is taking way too long (or is it suppose to?). This slows down anytime I use an integer past

Answer 1

first, use a long instead of an int, to avoid overflow.

Secondly, use a non-recursive algorithm, as a recursive one exists in exponential time I think. A well designed non-recursive one will solve in linear time (I think).

Example non-recursive

static long getSum(int n){

    long[] fibonacci = new long[n];
    fibonacci[0] = 1;
    fibonacci[1] = 1;

    if (n==0) return 0;
    if (n==1 || n==2) return 1;

    for(int i = 2; i < n;i++){
        fibonacci[i] = fibonacci[i-1]+ finonacci[i-2];
    }
    return fibonacci[n-1];
}

I haven't tested this, but it should work.

If you plan to call this method frequently, it might be prudent to store the array outside of the method, so that it is a simple lookup when doing this. This would provide a constant time solution for numbers that have already been calculated at least once. an example of that is below.

static long[] fibonacci= {1,1};
static long getSum(int n){

    if (n==0) return 0;
    if (n==1 || n==2) return 1;

    int old_length = fibonacci.length;

    if(fibonacci.length < (n-1)){
        fibonacci = Arrays.copyOf(fibonacci,n);
    }else{
        return fibonacci[n-1];
    }



    for(int i = old_length; i < n;i++){
        fibonacci[i] = fibonacci[i-1]+ finonacci[i-2];
    }
    return fibonacci[n-1];
}

Again, the example is untested, so a bit of debugging might be required.

Here is a linear time implementation of the algorithm that uses a constant overhead, instead of linear overhead.

static long getSum(int n){

    long currentNum = 0;
    long previousNum = 1;
    long previousNum2 = 1;

    if (n==0) return 0;
    if (n==1 || n==2) return 1;

    for(int i = 2; i < n;i++){
        currentNum = previousNum+ previousNum2;
        previousNum2 = previousNum;
        previousNum = currentNum;
    }
    return currentNum;
}