Dynamic Programing -- Leetcode problem 746. Min Cost Climbing Stairs

描述：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].

分析：简单的动态规划问题，要求找到上楼的最小花费，每次可以上一层楼或者两层。

思路一：这道题的动态规划很好解决，找到每次上一层或者两层的最小消耗就可以了，需要注意的是第一次上楼可以上一层或两层，因此第一二层的消耗为0。

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