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. - 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();//3vector<int> dp(n+1);//4dp[0]=dp[1]=0;for(int i=2;i<=n;i++)//3dp[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();//3vector<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)//3dp[i]=min(dp[i+1]+cost[i],dp[i+2]+cost[i]);//状态转移方程return min(dp[0],dp[1]);}
};