LintCode 366 Fibonacci

/* 1st method will lead to time limit */
/* the time complexity is exponential sicne T(n) = T(n-1) + T(n-2) */

class Solution {
    /**
     * @param n: an integer
     * @return an integer f(n)
     */
    public int fibonacci(int n) {
        // write your code here
        if (n == 1 || n == 2) {
            return (n-1);
        }

        // int sum = (n-1) + (n-2);

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

    }
}

/* 2nd method will need O(n) space, using DP */
/* T and S are both O(n) */

 public int fibonacci(int n) {
         // declare an array to store the result
         // it has to be n+2 to avoid out_of_bound
         int[] f = new int[n+2];
         f[1] = 0; // when input is 1 => zero
         f[2] = 1; // when input is 2 => 1

         int i = 3;
         while (i <= n) { // it has to be incremental instead of decremental
             f[i] = f[i-1] + f[i-2];
             i++;
         }

         return f[n];
     }

/* 3rd method will only need O(1) space */
/* We can optimize the space used in method 2 by storing the previous two numbers only */
/* because that is all we need to get the next Fibannaci number in series. */

 public int fibonacci(int n) {
     if (n < 3) return n-1;

     int first = 0;
     int second = 1;
     int third = 1;

     int i = 3;
     while (i <= n) {
         third = first + second;
         first = second;
         second = third;
         i++;
     }
     return third;
 }