[LeetCode]Min Cost Climbing Stairs
作者是 在线疯狂 发布于 2017年12月17日 在 LeetCode.

题目描述：

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].

题目大意：

有n级台阶，每次可以向上跳1至2级，上台阶的总花费为sum(cost[i])， i为所有踩过的台阶。

求最小花费。

解题思路：

动态规划（Dynamic Programming）

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

Python代码：

class Solution(object):
    def minCostClimbingStairs(self, cost):
        """
        :type cost: List[int]
        :rtype: int
        """
        size = len(cost)
        dp = [cost[0], cost[1]]
        for x in range(2, size):
            dp.append(min(dp[x - 1], dp[x - 2]) + cost[x])
        return min(dp[-1], dp[-2])

 

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