Min Cost Climbing Stairs

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本文探讨了在每个阶梯都有特定非负成本的情况下，寻找从地面到达楼顶的最小成本路径的问题。通过动态规划方法，分析了从第n步或n-1步达到顶层的最小成本，提供了一个C++实现的解决方案。

746. Min Cost Climbing Stairs

On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).

Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step with index 0, or the step with index 1.

Example 1:

Input: cost = [10, 15, 20]
Output: 15
Explanation: Cheapest is start on cost[1], pay that cost and go to the top.

Example 2:

Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
Output: 6
Explanation: Cheapest is start on cost[0], and only step on 1s, skipping cost[3].

分析：最后可以跳上top可以从min(f(n),f(n-1))来考虑。
对于f(n)一定是a[n]+min(f(n-1),f(n-2))来的。同时由于开始可能从0或者1起跳，所以对于开始几个点自己要分析下
f(4)=a[4]+min(f(3)+f(2))
f(3)=a[3]+min(f(1)+f(2))
f(2)=a[2]+min(f(1),f(0))
f(1)=a[1]
f(0)=a[0]

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        
        int costSize=cost.size();
        vector<int> fn(costSize);
        if(costSize==0)
            return 0;
        else if(costSize==1)
            return cost[0];
        else if(costSize==2)
            return cost[1];
        
        fn[0]=cost[0];
        fn[1]=cost[1];
 
        for(int i=2;i<costSize;i++)
        {
            fn[i]=cost[i]+min(fn[i-1],fn[i-2]);
            
            
            
        }
        return min(fn[costSize-1],fn[costSize-2]);
    }
};