[Leetcode 162]爬楼梯最小耗费Min Cost Climbing Stairs**

【题目】

爬楼梯变式，可以上1/2台阶，每个台阶有cost，求爬上去的最小耗费

You are given an integer array cost where cost[i] is the cost of ith step on a staircase. Once you pay the cost, you can either climb one or two steps.

You can either start from the step with index 0, or the step with index 1.

Return the minimum cost to reach the top of the floor.

 

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].

 

【思路】

• 全都能动态规划为爬一节/两节的问题
• 由初始的dp[0] dp[1]推导后面

【代码】

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        //说了length>=2
        int len=cost.length;
        int[] dp=new int[len];
        for(int i=0;i<len;i++){
            if(i<2)
                dp[i]=cost[i];
            else
                dp[i]=cost[i]+Math.min(dp[i-1],dp[i-2]);
        }
        return Math.min(dp[len-1],dp[len-2]);
    }
}