高效滚动最大和最小窗口 [英] Efficient Rolling Max and Min Window

问题描述

我想有效地计算滚动最大值和最小值.这意味着比每次窗口移动时从所有使用的值中重新计算最大值/最小值更好.

I want to calculate a rolling maximum and minimum value efficiently. Meaning anything better than recalculating the maximum/minimum from all the values in use every time the window moves.

这里有一篇帖子提出了同样的问题，有人发布了一个涉及某种堆栈方法的解决方案，该方法据称根据其评级有效.但是我这辈子都找不到它了.

There was a post on here that asked the same thing and someone posted a solution involving some kind of stack approach that supposedly worked based on its rating. However I can't find it again for the life of me.

在找到解决方案或帖子方面的任何帮助将不胜感激.谢谢大家！

Any help would be appreciated in finding a solution or the post. Thank you all!

推荐答案

您要使用的算法称为升序最小值 (C++ 实现).

The algorithm you want to use is called the ascending minima (C++ implementation).

要在 C# 中执行此操作，您需要获得一个 双端队列 类，并且NuGet 上有一个不错的应用，名为 Nito.Deque.

To do this in C#, you will want to get a double ended queue class, and a good one exists on NuGet under the name Nito.Deque.

我已经使用 Nito.Deque 编写了一个快速的 C# 实现，但我只是简单地检查了它，并且是从我的脑海中完成的，所以它可能是错误的！

I have written a quick C# implementation using Nito.Deque, but I have only briefly checked it, and did it from my head so it may be wrong!

public static class AscendingMinima
{
    private struct MinimaValue
    {
        public int RemoveIndex { get; set; }
        public double Value { get; set; }
    }

    public static double[] GetMin(this double[] input, int window)
    {
        var queue = new Deque<MinimaValue>();
        var result = new double[input.Length];

        for (int i = 0; i < input.Length; i++)
        {
            var val = input[i];

            // Note: in Nito.Deque, queue[0] is the front
            while (queue.Count > 0 && i >= queue[0].RemoveIndex)
                queue.RemoveFromFront();

            while (queue.Count > 0 && queue[queue.Count - 1].Value >= val)
                queue.RemoveFromBack();

            queue.AddToBack(new MinimaValue{RemoveIndex = i + window, Value = val });

            result[i] = queue[0].Value;
        }

        return result;
    }
}

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