关于此题我的往期文章：

leetCode 746. 使用最小花费爬楼梯 动态规划-CSDN博客https://heheda.blog.csdn.net/article/details/133325840

• dfs(i-1) 跳到 dfs(i) 需要花费 dfs(i-1) + cost[i-1]
• dfs(i-2) 跳到 dfs(i) 需要花费 dfs(i-2) + cost[i-2]

 （1） 递归（超时）

class Solution {
public:// 递归（超时）int minCostClimbingStairs(vector<int>& cost) {int n = cost.size();function<int(int)> dfs=[&](int i) -> int{if(i==0 || i==1) return 0;return min(dfs(i-1)+cost[i-1],dfs(i-2)+cost[i-2]);};return dfs(n);}
};

（2）递归搜索 + 保存计算结果 = 记忆化搜索

class Solution {
public:// 记忆化搜索int minCostClimbingStairs(vector<int>& cost) {int n = cost.size();vector<int> memo(n+1,-1);function<int(int)> dfs=[&](int i) -> int{if(i==0 || i==1) return 0;int& res = memo[i];if(res != -1) return res;return res = min(dfs(i-1)+cost[i-1],dfs(i-2)+cost[i-2]);};return dfs(n);}
};

（3）1:1 翻译成递推

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

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

• 空间优化

class Solution {
public: // 递推 + 空间优化int minCostClimbingStairs(vector<int>& cost) {int n = cost.size();vector<int> f(2,0);f[0]=0;f[1]=0;for(int i=2;i<=n;i++) {f[i%2] = min(f[(i-1)%2]+cost[i-1],f[(i-2)%2]+cost[i-2]);}return f[n%2];}
};

class Solution {
public:// 递推 + 空间优化int minCostClimbingStairs(vector<int>& cost) {int n = cost.size();int f0=0,f1=0;for(int i=2;i<=n;i++) {int tmp = min(f1+cost[i-1],f0+cost[i-2]);f0 = f1;f1 = tmp; }return f1;}
};

我的往期文章推荐：

LeetCode 70.爬楼梯 + 记忆化搜索 + 递推 + 动态规划 + 空间优化-CSDN博客https://blog.csdn.net/weixin_41987016/article/details/134192204?spm=1001.2014.3001.5501leetCode 746. 使用最小花费爬楼梯 动态规划-CSDN博客https://heheda.blog.csdn.net/article/details/133325840leetCode 198.打家劫舍 动态规划入门:从记忆化搜索到递推-CSDN博客https://blog.csdn.net/weixin_41987016/article/details/134179583?spm=1001.2014.3001.5501