tag：
- Easy;
- Dynamic Programming;
question：
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].
思路：
这道题是之前那道Climbing Stairs的拓展，这里不是求步数，而是每个台阶上都有一个cost，让我们求爬到顶端的最小cost是多少。换汤不换药，还是用动态规划DP来做。这里我们定义一个一维的dp数组，其中dp[i]表示爬到第i层的最小cost，然后我们来想dp[i]如何推导。我们来思考一下如何才能到第i层呢？是不是只有两种可能性，一个是从第i-2层上直接跳上来，一个是从第i-1层上跳上来。不会再有别的方法，所以我们的dp[i]只和前两层有关系，所以可以写做如下：
dp[i] = min(dp[i- 2] + cost[i - 2], dp[i - 1] + cost[i - 1])
最后我们返回最后一个数字dp[n]即可，参见代码如下：
class Solution {
public:
int minCostClimbingStairs(vector<int>& cost) {
int n = cost.size();
vector<int> dp(n+1, 0);
for (int i=2; i<n+1; ++i) {
dp[i] = min(dp[i-2]+cost[i-2], dp[i-1]+cost[i-1]);
}
return dp.back();
}
};
