. - 力扣（LeetCode）. - 备战技术面试？力扣提供海量技术面试资源，帮助你高效提升编程技能,轻松拿下世界 IT 名企 Dream Offer。https://leetcode.cn/problems/min-cost-climbing-stairs/description/从左向右填dp表

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n=cost.size();//3
        vector<int> dp(n+1);//4
        dp[0]=dp[1]=0;
        for(int i=2;i<=n;i++)//3
            dp[i]=min(dp[i-1]+cost[i-1],dp[i-2]+cost[i-2]);
        return dp[n];
    }
};

从右向左填dp表

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n=cost.size();//3
        vector<int> dp(n);//4 
        //初始化
        dp[n-1]=cost[n-1],dp[n-2]=cost[n-2];
        for(int i=n-3;i>=0;--i)//3
            dp[i]=min(dp[i+1]+cost[i],dp[i+2]+cost[i]);//状态转移方程
        return min(dp[0],dp[1]);
    }
};