746. Min Cost Climbing Stairs (easy)

Q:

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].

A:

这个的主要的坑，就是一开始没有注意到， 最后一节也是要爬的，因为，目标是走上floor

-------------简单的 DP -----

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        // at the last step, we need also climb,  fuck
        int n = cost.size();
        vector<int> vec(n+2, INT_MAX);  //extra as floor, as we need move last step
        vec[0] = 0;
        vec[1] = 0;
        // for each step, calculate the ones that it can reach        
        for(int i =0;i<n;i++)
        {
            vec[i+1] = min(vec[i+1], cost[i]+vec[i]);
            vec[i+2] = min(vec[i+2], cost[i]+vec[i]);
        }
        return vec[n];
    }
};