In this Leetcode Fibonacci Number problem solution The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,

F(0) = 0, F(1) = 1

F(n) = F(n - 1) + F(n - 2), for n > 1.

Given n, calculate F(n).

Leetcode Fibonacci Number problem solution


Problem solution in Python.

class Solution:
    def fib(self, N: int) -> int:
        if N == 0:
            return 0
        if N == 1:
            return 1
        curr_n = 2
        n_1 = 1
        n_2 = 0
        
        while N>=curr_n:
            ans = n_1 + n_2
            n_2=n_1
            n_1=ans
            curr_n+=1
        return ans

Problem solution in Java.

public int fib(int N) {
    if(N == 1) return 1;
    if(N == 0) return 0;
    return fib(N-1)+fib(N-2);
}


Problem solution in C++.

public:
    int fib(int N) {
        if(N < 2){
            return N;
        }
        vector vals = {0, 1};
        
        int i = 2;
        int sum = 1;
        while(i <= N){
            sum = vals[i-1] + vals[i-2];
            vals.push_back(sum);
            i++;
        }
        
        return sum;
    }
};


Problem solution in C.

int calculate_fib(int N, int *cache)
{
    if (N == 1 || N == 0)
        return (N);
    if (cache[N - 1] != 0)
        return (cache[N - 1]);
    cache[N - 1] = calculate_fib(N - 1, cache) + calculate_fib(N - 2, cache);
    return (cache[N - 1]);
}

int fib(int N){
    int *cache;
    int r;
    int i;
    
    i = -1;
    cache = (int *)malloc(sizeof(int) * N);
    while (++i < N)
        cache[i] = 0;
    r = calculate_fib(N, cache);
    free(cache);
    return (r);
}