70. Climbing Stairs

最新推荐文章于 2024-01-18 13:59:59 发布

转载最新推荐文章于 2024-01-18 13:59:59 发布 · 42 阅读

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原文链接：http://www.cnblogs.com/neweracoding/p/5390795.html

本文探讨了经典的爬楼梯问题，提供了两种基于动态规划的解决方案。第一种使用递归加缓存的方法来减少重复计算，第二种采用迭代的方式更新状态数组，最终返回到达顶部的不同路径数量。

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?

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Dynamic Programming

Solution 1:

public class Solution {
    public int climbStairs(int n) {
        if(n <=2 )
            return n;
        
        int[] count = new int[n];
        count[0] = 1;
        count[1] = 2;
        
        return getCount(count, n-1);
    }
    
    private int getCount(int[] count, int index)
    {
        if(count[index] != 0)
            return count[index];
        int steps = getCount(count, index -1) + getCount(count, index -2);
        count[index] = steps;
        return steps;
    }
}

Solution 2:

public class Solution {
    public int climbStairs(int n) {
        if(n <=2 )
            return n;
        
        int[] count = new int[n];
        count[0] = 1;
        count[1] = 2;
        
        for(int i =2; i< n; ++i)
        {
            count[i] = count[i-1]+count[i-2];
        }
        
        return count[n -1];
    }
}

转载于:https://www.cnblogs.com/neweracoding/p/5390795.html