746. Min Cost Climbing Stairs(动态规划）

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].

分析：
这道题比较简单了，是斐波那契模型的变体，dp[i]表示调到第i阶的最小cost，很容易得到dp[i]=min(dp[i-1]+cost[i],dp[i-2]+cost[i-2]);

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int dp[cost.size()+1]={0};
        for(int i=2;i<cost.size()+1;i++)
            dp[i]=min(dp[i-1]+cost[i-1],dp[i-2]+cost[i-2]);
        return dp[cost.size()];
    }
};