746. Min Cost Climbing stairs

Problem

On a staircase, cost[i] represents the cost of the I-ladder, as long as you pay the price you can climb one or two steps, you can start from index 0 or index 1 to crawl. The minimum price to climb to the top (the cost index from 0 to n-1, to climb to the ladder on Index N). The cost is at least 2 of the length.

Ideas

Procedure 1: Dp[i] represents the minimum cost to reach the I ladder, as one or two steps can be skipped, dp[i] can represent the minimum value of the cost of jumping over the previous ladder and jumping over the first two steps, the DP formula is:\ (Dp[i] = min (dp[ I-1]+cos[i-1], dp[i-2]+cost[i-2]) \)

Time complexity O (n), spatial complexity O (n)

Procedure 2: Considering that the value of each DP Dp[i] depends only on the previous two values dp[i-1] and dp[i-2], you can use two temporary variables F1 and F2 instead, which saves the space overhead of the DP array.

Time complexity O (n), Spatial complexity O (1)

Code

Procedure 1:

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

Procedure 2:

class Solution(object):    def minCostClimbingStairs(self, cost):        """        :type cost: List[int]        :rtype: int        """        f1 = f2 = 0        for i in range(2,len(cost)+1):            f1, f2 = min(f1+cost[i-1], f2+cost[i-2]), f1        return f1   

746. Min Cost Climbing stairs