Leetcode 48. Min Cost Climbing Stairs

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本文介绍了一个经典的动态规划问题——爬楼梯。目标是从两种起步方式中选择路径，通过支付阶梯费用到达楼顶，使得总花费最小。文章提供了详细的解题思路及C++实现代码。

48. 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].

Note:

cost will have a length in the range [2, 1000].
Every cost[i] will be an integer in the range [0, 999].

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

题目大意：爬阶梯，可以从下标为0或1的阶梯开始，每次爬一步或两步，每爬一次需要支付对应阶梯的费用，cost数组存放每级阶梯的对应费用，直到爬完所有阶梯，求支付的最少费用为多少。

解题思路：dp递推，爬当前阶梯支付的最少费用为：当前阶梯的费用＋ min（前一级阶梯的最少费用，前前一级阶梯的最少费用），递推式为：

dp[i] = min(dp[i-2] + cost[i],dp[i-1] + cost[i]);

代码如下：

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n = cost.size();
        int inf = (1 << 31) - 1 ;
        if(n == 0)
            return 0;
        int dp[n + 5];
        for(int i = 0;i < n + 5;i ++)
            dp[i] = inf;
        dp[0] = cost[0];
        dp[1] = cost[1];
        for(int i =2;i < n;i ++){
            dp[i] = min(dp[i-2] + cost[i],dp[i-1] + cost[i]);
        }
        return min(dp[n-1],dp[n-2] );
    }
};