问爬楼梯的最小费用动态规划

EN

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.

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        return helper(cost, cost.size() - 1);
    }

    int helper(vector<int>& cost, int currStair) {
        static vector<double> minCost(cost.size(), 0);
        minCost[0] = cost[0];
        minCost[1] = cost[1];
        if (minCost[currStair] > 0) {
            return minCost[currStair];
        }

        return minCost[currStair] = min(helper(cost, currStair - 1), helper(cost, currStair - 2)) + cost[currStair];
    }
};

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        return helper(cost, cost.size() - 1);
    }

    int helper(vector<int>& cost, int currStair) {
        if (currStair < 0 || cost.size() <= 1) {
            return 0;
        }

        static vector<double> minCost(cost.size(), 0);
        minCost[0] = cost[0];
        minCost[1] = cost[1];

        if (minCost[currStair] > 0) {
            return minCost[currStair];
        }

        minCost[currStair] = min(helper(cost, currStair - 1), helper(cost, currStair - 2)) + cost[currStair];

        return min(minCost[cost.size()-1], minCost[cost.size()-2]);
    }
};