Climbing Stairs

问题描述

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?

算法

有n阶，第一次迈一阶，剩余的跟n-1的次数相同，或者第一次迈2阶，剩余的跟n-2的次数相同。斐波那契数列。

代码一：

public int climbStairs(int n) {
        int[] a=new int[n+1];
        if(n<3)
            return n;
        else{
            a[1]=1;
            a[2]=2;
            for(int i=3;i<=n;i++){
                a[i]=a[i-1]+a[i-2];
            }
            return a[n];
        }  
    }

代码二

public int climbStairs(int n){
        if(n<3)
            return n;
        else{
            int beforeOne=2,beforeTwo=1,sum = 0;
            for(int i=3;i<=n;i++){
                sum=beforeOne+beforeTwo;
                beforeTwo=beforeOne;
                beforeOne=sum;
            }
            return sum;
        }
    }

注意事项

下面的代码会超时

public class Solution {
    public int climbStairs(int n) {
        if(n==1||n==2)
            return n;
        else
            return climbStairs(n-1)+climbStairs(n-2);
    }
}