如何找到范圍的最大值和最小值

Question

代碼正在運行，但我想知道如何在“突破柱”出現后找到范圍內最后三個柱的最高值和最低值。

在下面的示例中，該值為 2。

protected override void OnBarUpdate()
{
    if (CurrentBar < 3)
    {
        Value[0] = 0;
    }
    else if((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Close[0] > High[1]))
    {
        Value[0] = 2;           
    }
    else if ((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Close[0] < Low[1]))
    {
        Value[0] = -2;
    } else if ((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Close[0] > Close[1]))
    {
        Value[0] = 1;
    } else if ((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Close[0] < Close[1]))
    {
        Value[0] = -1;
    }
    else
    {
        Value[0] = 0;
    }
}

Answer 1

我按照@kirodge 的建議使用了 MAX 和 MIN function 並且效果很好。 這是解決方案。

protected override void OnBarUpdate()
{
    {
    if (CurrentBar < 3)
    {
        Value[0] = 0;
    }
    else if((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Low[0] < Low[3] && Close[0] > MAX(High, 3)[1]))
    {
        Value[0] = 2;

    }
    else if ((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Low[0] < Low[3] && Close[0] < MIN(Low, 3)[1]))
    {
        Value[0] = -2;
    } else if ((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Low[0] < Low[3] && Close[0] > Open[1]))
    {
        Value[0] = 1;
    } else if ((High[0] > High[1] && High[0] > High[2] && High[0] > High[3] && Low[0] < Low[1] && Low[0] < Low[2] && Low[0] < Low[3] && Close[0] < Open[1]))
    {
        Value[0] = -1;
    }
        else {
        Value[0] = 0;
    }
}
}