Problem #12 [Hard]-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_51444711/article/details/137957619

文章讨论了如何使用动态规划算法解决阶梯问题，给出了两种情况：一种是每次只能爬1或2级台阶，另一种是可以爬取给定的一组正整数。通过计算到达第n层楼梯的不同路径数，展示了递推公式dp[i]=dp[i-1]+dp[i-2]的应用。

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This problem was asked by Amazon.

There exists a staircase with N steps, and you can climb up either 1 or 2 steps at a time. Given N, write a function that returns the number of unique ways you can climb the staircase. The order of the steps matters.

For example, if N is 4, then there are 5 unique ways:

1, 1, 1, 1
2, 1, 1
1, 2, 1
1, 1, 2
2, 2
What if, instead of being able to climb 1 or 2 steps at a time, you could climb any number from a set of positive integers X? For example, if X = {1, 3, 5}, you could climb 1, 3, or 5 steps at a time.
典型的动态规划。首先确定dp数组定义，即爬i层楼梯有dp[i]种方法；然后思考递推公式dp[i] = dp[i-1] + dp[i-2]。其中dp[1]=1, dp[2]=2。

class Solution {
public:
    int climbStairs(int n) {
        if (n <= 1) return n; 
        vector<int> dp(n + 1);
        dp[1] = 1;
        dp[2] = 2;
        for (int i = 3; i <= n; i++) { // 注意i是从3开始的
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }
};

对于follow-up的问题，递推公式稍稍变化即可。

class Solution {
public:
    int climbStairs(int n, vector<int> X) {
        if (n <= 1) return n; 
        vector<int> dp(n + 1);
        dp[0] = 1;
        for (int i = 1; i <= n; i++) { // 注意i是从2开始的
        	for(int j:X){
        		if(i>=j) dp[i] += dp[i-j]
        	}  
        }
        return dp[n];
    }
};