Description：
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].

题意：给定一个一维数组，元素xi表示第i个台阶需要花费的代价（台阶下标从0开始）；现要求计算花费最小的大家到达台阶的顶部；

解法：这道题目我们可以采用动态规划的思想，我们定义数组minCost表示处在第i个台阶时所要花费的最小代缴，那么满足下列等式

minCost[i]=cost[i]+min(minCost[i−1],minCost[i−2]) m i n C o s t [ i ] = c o s t [ i ] + m i n ( m i n C o s t [ i − 1 ] , m i n C o s t [ i − 2 ] )

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