[Leetcode] 746. Min Cost Climbing Stairs 解题报告

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]

.
思路：

这是最简单和最原始的动态规划问题了。我们定义dp[i]表示从第i个位置上开始出发需要付出的最小代价，那么递推公式为：dp[i] = cost[i] + min(dp[i - 2], dp[i - 1])。最后我们返回dp数组中倒数两个元素中的小者即可。

代码：

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