【12月20日】LeetCode刷题日志（四）：Min Cost Climbing Stairs

最新推荐文章于 2021-04-10 14:46:54 发布

原创最新推荐文章于 2021-04-10 14:46:54 发布 · 350 阅读

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本文介绍了一种寻找爬楼梯最低成本的算法实现方案。通过动态规划的方法，计算从楼梯底部到顶部所需的最低费用，允许每次跳跃一阶或两阶。

题目描述
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.

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].

分析
这是一个时序的求最小值的问题，此时的状态依赖于前一时的状态。令M[i]表示到达第i各台阶之前的min，则M[0] = M[1] =0，且对于i>=2,有 M[i] =Math.min(M[i-2]+cost[i-2], M[i-1]+cost[i-1]).这是分析的关键一步，有了以上的分析，最终的结果只要return Math.min(M[len-1]+cost[len-1], M[len-2]+cost[len-2])即可。

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

    public static void main(String[] args) {
        Solution solution = new Solution();
        int[] cost = new int[]{1, 100, 1, 1, 1, 100, 1, 1, 100, 1};
        int result = solution.minCostClimbingStairs(cost);
        System.out.println(result);
    }
}