LeetCode - Climbing Stairs | Darren's Blog

You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

By taking one step, there are T(n−1) distinct ways to finish the remaining. And by taking two steps, there are T(n−2) distinct ways to finish the remaining. That is, a recursion relationship can be identified:

T(n)=T(n−1)+T(n−2).

This is exactly the definition of Fibonacci sequence.

public int climbStairs(int n) {

        int f1 = 1, f2 = 2;

        if (n == 1)

            return f1;

        if (n == 2)

            return f2;

        for (int i = 3; i <= n; i++) {

            int f3 = f1 + f2;

            f1 = f2;

            f2 = f3;

        }

        return f2;

    }

Another possible way is to leverage its matrix representation:

(Fn+1FnFnFn−1)=(1110)n+1.

可以转换成斐波那契数列进行解答
Code from http://my.oschina.net/u/811399/blog/141550
public class Solution {
    public int climbStairs(int n) {
        // Start typing your Java solution below
        // DO NOT write main()
    double s = Math.sqrt(5);
   return (int)Math.floor(1/s*( Math.pow(0.5+s/2,(double)n+1)-Math.pow(0.5-s/2,(double)
    }
}
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