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].

答案
Let mincost[i] be the total cost in order to get to stair i.
We want to know mincost[n]

Base cases
mincost[0] = 0
mincost[1] = 0

Recursive cases
mincost[i] = min(mincost[i - 1] + cost[i-1], mincost[i - 2] + cost[i - 2])

Implementation

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int[] mincost = new int [cost.length + 1];
        for(int i = 2; i <= cost.length; i++)
            mincost[i] = Math.min(mincost[i - 1] + cost[i - 1], mincost[i - 2] + cost[i - 2]);
        return mincost[cost.length];
    }
}

如果仔细想一下，会发现以上答案中每次循环计算mincost[i]时，只用到了mincost[i-1]和mincost[i-2], 而 mincost[i-3]以及之前的数都是闲置的，所以完全可以只用2个变量来代替mincost数组

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int mincost1 = 0, mincost2 = 0, mincost3 = 0; 
        for(int i = 2; i <= cost.length; i++) {
            mincost3 = Math.min(mincost1 + cost[i - 1], mincost2 + cost[i - 2]);
            mincost2 = mincost1;
            mincost1 = mincost3;
        }
        return mincost3;
    }
}