leetcode 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].

Note:

1. cost will have a length in the range [2, 1000].
2. Every cost[i] will be an integer in the range [0, 999].

class Solution(object):
    def minCostClimbingStairs(self, cost):
        """
        :type cost: List[int]
        :rtype: int
        """
        # mincost = min(mincost(n-1), mincost(n-2))+cost[n]
        # n = 2               
        a = cost[0]
        b = cost[1]
        for i in range(2, len(cost)):
            b,a = min(a, b)+cost[i], b
        return min(b, a)

注意：本质上是dp，dp[i]表示经过step i的min cost。

那么最后一步cost应该是min(dp[i], dp[i-1]) 表示要么最后一步是踩step i，cost就是dp[i]，要么不踩step[i]，必然是从step i-1过来的，跨了两步。

空间O（n）的解法：

Solution #1: Bottom-Up dynamic programming

Let dp[i] be the minimum cost to reach the i-th stair.

Base cases:

dp[0]=cost[0]
dp[1]=cost[1]

DP formula:

dp[i]=cost[i]+min(dp[i-1],dp[i-2])

Note: the top floor n can be reached from either 1 or 2 stairs away, return the minimum.

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

或者是：

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int [] mc = new int[cost.length + 1];
        mc[0] = cost[0];
        mc[1] = cost[1];
        
        for(int i = 2; i <= cost.length; i++){
            int costV = (i==cost.length)?0:cost[i];
            mc[i] = Math.min(mc[i-1] + costV, mc[i-2] + costV);
        }
        return mc[cost.length];
    }
}