[LeetCode] 746. Min Cost Climbing Stairs

最新推荐文章于 2025-08-19 16:19:29 发布

转载最新推荐文章于 2025-08-19 16:19:29 发布 · 78 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/lightwindy/p/9646839.html

文章标签：

#数据结构与算法 #java #python

本文探讨了在每一级阶梯都有一定成本的情况下，寻找从底部到顶部的最小成本路径的算法。通过动态规划（DP）的方法，文章提供了多种语言的实现方案，包括Java、Python和C++，并对比了不同实现方式的优劣。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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].

解法：DP，一种是设dp长度为n+1, 公式：dp[i] = min(dp[i- 2] + cost[i - 2], dp[i - 1] + cost[i - 1])，最后返回dp的最后一个值。

另一种是设dp长度为n, 公式：dp[i] = cost[i] + min(dp[i- 1], dp[i - 2])，最后返回倒数的两个数组值中小的一个。

Java: dp + greedy

public int minCostClimbingStairs(int[] cost) {
        for (int i = 2; i < cost.length; i++) {
            cost[i] += Math.min(cost[i-1], cost[i-2]);
        }
        return Math.min(cost[cost.length-1], cost[cost.length-2]);
    }

Java:

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int [] mc = new int[cost.length + 1];
        mc[0] = cost[0];
        mc[1] = cost[1];
        
        for(int i = 2; i <= cost.length; i++){
            int costV = (i==cost.length)?0:cost[i];
            mc[i] = Math.min(mc[i-1] + costV, mc[i-2] + costV);
        }
        return mc[cost.length];
    }
}

Python:

class Solution(object):
    def minCostClimbingStairs(self, cost):
        """
        :type cost: List[int]
        :rtype: int
        """
        dp = [0] * 3
        for i in reversed(xrange(len(cost))):
            dp[i%3] = cost[i] + min(dp[(i+1)%3], dp[(i+2)%3])
        return min(dp[0], dp[1])

Python:

def minCostClimbingStairs(self, cost):
    n = len(cost)
    if n == 0 or n == 1:
        return 0
    min_cost0, min_cost1 = cost[0], cost[1]
    for i in range(2, n):
        min_cost0, min_cost1 = min_cost1, min(min_cost0, min_cost1) + cost[i]

    return min(min_cost0, min_cost1)

C++:

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

C++:

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

类似题目：

[LeetCode] 70. Climbing Stairs 爬楼梯

[LeetCode] 53. Maximum Subarray 最大子数组

[Airbnb] Max Sum of Non-consecutive Array Elements