Climbing Stairs

最新推荐文章于 2021-11-04 20:26:38 发布

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本文探讨了经典的爬楼梯问题，通过动态规划的方法来找出达到楼梯顶部的不同方式的数量。该问题要求每次只能上1阶或2阶，求总的上楼方案数。

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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?

Analysis: It's a DP problem. For each n, the last step can either climb 1 or 2 steps. So, f(n) = f(n-1) + f(n-2)

public class Solution {
    public int climbStairs(int n) {
        // the last step can climb either 1 or 2 steps, so ways = climbStairs(n-1) + climbStairs(n-2)
        if(n==0) return 0;
        else if(n == 1) return 1;
        
        int [] ways = new int[n+1];
        ways[1] = 1;
        ways[2] = 2;
        for(int i=3; i<ways.length; i++) {
            ways[i] = ways[i-1] + ways[i-2];
        }
        return ways[n];
    }
}