746. Min Cost Climbing Stairs

　　　　这道题为简单题

　　题目：

 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:

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

 

　　思路：

　　　　这个题用DP还是比较简单的，主要找到状态转换方程就比较容易了。用列表a来定义每一步的最优答案，从而找到转移方程 a[i] = min(a[i-2] + cost[i], a[i-1] + cost[i])

　　代码：

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