Climb Stairs - dynamic programming

Question

from lintcode
You are climbing a staircase. It takes n steps to reach the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
Example
Given an example n=3 , 1+1+1=2+1=1+2=3
return 3

Ideas

see code comments;

/**
 *  Formula:
 *  number of paths to arrive Xth stair =
 *      number of paths to arrive X - 1 th stair
 *      +
 *      number of paths to arrive X - 2 th stair
 *
 *  That is:
 *  f(x) = f(x - 1) + f(x - 2)
 *
 *  You need to remember 2 states: x - 1 and x - 2 cases.
 *
 */
public class Solution {

    private int numOfStates = 2;

    private int indexOf(int n) {
        return n % numOfStates;
    }

    /**
     * @param n: An integer
     * @return: An integer
     */
    public int climbStairs(int n) {

        /**
         * edge case
         */
        if (n == 0) return 0;

        int[] total = new int[numOfStates];

        /**
         *  Initial states
         */
        total[indexOf(1)] = 1;
        total[indexOf(2)] = 2;

        /**
         * calculation: f(x) = f(x - 1) + f(x - 2)
         */
        for(int x = 3; x <= n; x++)
            total[indexOf(x)] =
                    total[indexOf(x - 1)] + total[indexOf(x - 2)];

        // calculated result after dynamic programming
        return total[indexOf(n)];
    }
}