70. Climbing Stairs

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?

Note:Given n will be a positive integer.

有n阶楼梯，一次可以上一阶、也可以上2阶，求总共有多少种不同的走法

这是斐波那契数列问题,其中一个有名的问题：兔子繁殖。如下图，

按照上面的思路，给出代码：

public int climbStairs(int n) {
        if(n <= 0 ) return 0;
        if(n == 1) return 1;
        if(n == 2) return 2;
        int sum = 0;
        int n_2 = 1; //A(n-2)=1
        int n_1 = 2;  // n-1=2
        for(int i = 2; i < n; i++)
        {
            sum = n_1 + n_2;
            n_2 = n_1;
            n_1 = sum;
        }
        return sum;
    }

每一阶段有多少中走法，可以使用一个数组记录下来

先初始化要给全为0的数组

int a[] = new int[n]; //n阶楼梯，eg,a[4]表示5阶楼梯的走法

public static int climbStairs(int a[],int m)
    {
        a[0] = 1;
        a[1] = 2;
        for(int i = 2;i <= m; i++)
            a[i] = a[i - 1] + a[i - 2];
        return a[m] ;
    }