LeetCode 746. Min Cost Climbing Stairs

LeetCode 746. Min Cost Climbing Stairs

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].

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分析：

动态规划的问题：
假设dp[i]表示到达第i阶楼梯的最小代价
初始化dp[0]=cost[0], dp[1]=cost[1]
状态转移方程：
dp[i] = cost[i] + min(dp[i-1] + dp[i-2])
结果：因为最后一步可以是跨一阶或者二阶楼梯，故返回min(dp[n-1] + dp[n-2])。

Example 2分析：

从0开始计算，即0表示第1阶楼梯，9表示第10阶楼梯：
dp[0] = 1
dp[1] = 100
dp[2] = 1 + 1 = 2
dp[3] = 1 + 2 = 3
dp[4] = 1 + 2 = 3
dp[5] = 100 + 3 = 103
dp[6] = 1 + 3 = 4
dp[7] = 1 + 4 = 5
dp[8] = 100 + 4 = 104
dp[9] = 1 + 5 = 6

最终爬楼梯的方式是这样的：0->2->4->6->7->9.

代码如下：

#include 
#include 
#include 
using namespace std;
class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n = cost.size();
        vector<int> dp(n);
        dp[0] = cost[0];
        dp[1] = cost[1];
        for (int i = 2; i < n; i++) {
            dp[i] = cost[i] + min(dp[i - 2], dp[i - 1]);
        }
        return min(dp[n - 2], dp[n - 1]);
    }
};
int main() {
    int n;
    cin >> n;
    vector<int> cost(n);
    for (int i = 0; i < n; i++) {
        cin >> cost[i];
    }
    Solution s;
    cout << s.minCostClimbingStairs(cost) << endl;
    return 0;
}