数组中的位置(偏移量)c#

Position in arrays (offset) c#

我遇到了无法有效解决的问题。 我需要做的是: 我在数组中有一个起始位置(在我的例子中是列表),我也有一个偏移量。 int类型的偏移量:

当偏移量 > 0 时,我用这个来计算新位置:

        if (currentPosition + offset < lenght)
        {
            return currentPosition + offset;
        }
        return (currentPosition + offset)%lenght;

问题是当偏移量 < 0 时:

        for (int i = 0; i < offset * -1; i++)
        {
            currentPosition -= 1;
            if (currentPosition == -1)
            {
                currentPosition = lenght - 1;
            }
        }
        return currentPosition;

但是这个解决方案真的很慢。 你们有什么想法吗? 提前谢谢你。

看起来 currentPosition 是一个整数。所以你可以只做计算,如果小于零则在之后更正;

currentPosition = (currentPosition + offset) % lenght;
if (currentPosition<0)
    currentPosition += lenght;
return currentPosition;

我想出了这个功能,希望它能有所帮助(为清楚起见添加了 in-code 评论):

private int CalcNewPosition(int[] arr, int position, int offset)
{
    if (position < 0 || position >= arr.Length)
        throw new ArgumentOutOfRangeException("position");

    // Calculate correct offset that is within bounds of array
    // by using modulus of offset divided by array length.
    var offsetOk = offset % arr.Length;

    // If offset is negative, calculate how many steps to
    // move forward instead of backwards.
    if (offsetOk < 0)
    {
        offsetOk = arr.Length + offsetOk;
    }

    // Calculate new offset
    var result = position + offsetOk;

    // If offset is greater or equal than length of array
    // set it to number of elements from beginning by
    // calculating the difference between length and new offset
    if (result >= arr.Length)
    {
        result = result - arr.Length;
    }

    return result;
}

我已经用这个调用试过了,它们都工作正常(我希望):

var pos1 = CalcNewPosition(arr, 3, 2);
var pos2 = CalcNewPosition(arr, 3, -1);
var pos3 = CalcNewPosition(arr, 3, -56);

希望对您有所帮助。

给出

(A) 0 < length && length <= int.MaxValue / 3
(B) 0 <= position && position < length
(C) -length < offset && offset < length

计算可以是

position = (position + offset + length) % length;

如果(C)不成立,我们可以把它变成与offset % length相同的情况,公式将改为

position = (position + (offset % length) + length) % length;