topic:
Idea:
According to the meaning of the question, we first clarify the meaning of dp array i. It is obvious here that i is the minimum cost of the corresponding ladder,
Secondly, in dp initialization, our dp[0] and dp[1] are 0 costs,
This is what we can choose. When dp[2] is reached, it is our min(dp[0] + cost[0],dp[1] + cost[1])
That is, this is our recursive formula: the minimum cost to reach the current ladder + the cost of the current ladder = the target ladder reached
即 dp[i] = min(dp[i - 1] + cost[i - 1],dp[i - 2] + cost[i - 2])；
The code is explained in detail below:
int minCostClimbingStairs(vector<int>& cost)
{
// 计算台阶数量
int n =cost.size();
// 定义 dp 数组，其中 dp[i]
// i 所对应的是相应台阶的最少花费
vector<int>dp(n + 1,0);
// dp 数组初始化，由于可以选择在 0 或 1 台阶开始爬楼梯
// 所以 先计算第三个台阶的最少花费
dp[2] = min(cost[0],cost[1]);
for(int i = 3;i <= n;++i)
{
// 递推公式，达到当前阶梯的最少花费 + 当前阶梯需要的花费 = 到达的目标阶梯
dp[i] = min(dp[i - 1] + cost[i - 1],dp[i - 2] + cost[i - 2]);
}
/*
打印 dp 数组查看是否是自己需要的效果
验证答案
debugv(dp);
*/
// 输出对应阶梯的最少花费
return dp[n];
}