# 746. Min Cost Climbing Stairs 使用最小花费爬楼梯

@TOC

## # 题目描述

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].

## # 解题方法

### # 动态规划

``````class Solution(object):
def minCostClimbingStairs(self, cost):
"""
:type cost: List[int]
:rtype: int
"""
costed = [0, 0]
for i in xrange(2, len(cost)):
costed.append(min(costed[i - 1] + cost[i - 1], costed[i - 2] + cost[i - 2]))
return min(costed[-1] + cost[-1], costed[-2] + cost[-2])
``````

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

## # 日期

2018 年 1 月 28 日 2018 年 11 月 17 日 —— 美妙的周末，美丽的天气