[LeetCode] 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:

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

上楼梯算法的花费。

规定每次可以跨1或者2节台阶。你可以任意从第一节或者第二节开始。添加的额外条件是当跨到每个台阶时，需要当前台阶的花费，求登上最高层台阶的最少花费。

这也就是一个动态规划问题。

首先创建dp数组，找出约束条件，统计2种登上台阶的方式所需要的花费的最小值。存入dp数组，即dp[i] = min(cost[i-2] + dp[i-2], cost[i-1] + dp[i-1]) 最后一个dp[n]就是所求的最短花费。参考代码如下

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        vector<int> dp(cost.size() + 1, 0);
        for (int i = 2; i < dp.size(); i++) {
            dp[i] = min(dp[i - 2] + cost[i - 2], dp[i -  1] + cost[i - 1]);
        }
        return dp.back();
    }
};
// 12 ms

也可以通过递归的方法求解。

t : 如果到当前台阶的最小花费

a: 到当前台阶前一次所需要的最小花费

b: 维护当前台阶的最小花费。

min(a, b): 完成爬楼梯操作后的最小花费。

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int a = 0, b = 0;
        for (auto c : cost) {
            int t = min(a, b) + c;
            a = b;
            b = t;
        }
        return min(a, b);
    }
};
// 9 ms

 

转载于:https://www.cnblogs.com/immjc/p/8073418.html