746. Min Cost Climbing Stairs ---E

You are given an integer array cost where cost[i] is the cost of ith step on a staircase. Once you pay the cost, you can either climb one or two steps.

You can either start from the step with index 0, or the step with index 1.

Return the minimum cost to reach the top of the floor.

 

Example 1:

Input: cost = [10,15,20]
Output: 15
Explanation: You will start at index 1.
- Pay 15 and climb two steps to reach the top.
The total cost is 15.

Example 2:

Input: cost = [1,100,1,1,1,100,1,1,100,1]
Output: 6
Explanation: You will start at index 0.
- Pay 1 and climb two steps to reach index 2.
- Pay 1 and climb two steps to reach index 4.
- Pay 1 and climb two steps to reach index 6.
- Pay 1 and climb one step to reach index 7.
- Pay 1 and climb two steps to reach index 9.
- Pay 1 and climb one step to reach the top.
The total cost is 6.

 

Constraints:

• 2 <= cost.length <= 1000
• 0 <= cost[i] <= 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];
    }
};

------------------------第二遍----------------

class Solution {

public:

int minCostClimbingStairs(vector<int>& cost) {

int n = cost.size();

int pre=0, prepre = 0;

for(int i =0; i<n;i++){

int cur = cost[i] + min(pre, prepre);

prepre = pre;

pre = cur;

}

return min(pre, prepre);

}

};