【LeetCode】746. Min Cost Climbing Stairs

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].

链接：https://leetcode.com/problems/min-cost-climbing-stairs/description/

题解：动态规划，用递归很简单， 主要有向前和向后，f[i]=cost[i]+min(f[i-1],f[i-2]),f[i]=cost[i]+min(f[i+1],f[i+2]);

代码：

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
         int f1=0,f2=0;
         for(int i=cost.size()-1;i>=0;i--){
            int f0=min(f1,f2)+cost[i];
            f2=f1;
            f1=f0;
         }
         return min(f1,f2);
    }
    int minCostClimbingStairs(vector<int>& cost) {
         int f1=0,f2=0;
         for(int i=0;i<cost.size();i++){
            int f0=min(f1,f2)+cost[i];
            f2=f1;
            f1=f0;
         }
         return min(f1,f2);
    }
    
};