如何从数组中找到最远的3个连续元素

Question

我有一系列可能的位置和另一个填充位置数组,这是一个子阵列possiblePositionsArray.possiblePositionsArray是固定的,已经确定.我想在所选数组元素的x位置的右侧和左侧找到最远的3个连续点filledPositions.让我用这个例子进一步解释.说

possiblePositionsArray = [p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15]
filledPositions = [p1, p2, p4, p7, p8, p9, p10, p12, p13, p14, p15]

两者都是CGPoints的数组,并且都具有相同的y位置并且按升序排列.如果我选择p11.x以下将是右侧和左侧的3个连续点.

[p7, p8, p9] and [p8, p9, p10] To the Left of p11
[p12, p13, p14] and [p13, p14, p15] to the right of p11

但是左边和右边最远的是:

farthest to left of p11 is [p7, p8, p9]
farthest to right of p11 is [p13, p14, p15]

我怎样才能做到这一点？

Answer 1

首先从头开始filledPositions。filledPositions查找中的第一项possiblePositionsArray。检查两个数组中接下来的两项是否彼此匹配。第一个连续组位于所选元素左侧最远的位置。即使元素x中的值possiblePositionsArray间隔不等距，这种方法也有效。

之后，您以相反的顺序执行此操作，找到最右边的。

其代码如下：

let selectedElement = yourSelectedElement

//left consecutive group   
var consLeft = [CGPoint]()
//right consecutive group
var consRight = [CGPoint]()

if filledPositions.count >= 3 {
    for i in 0..<filledPositions.count-2 {
        // find the index of the element from filledPositions in possiblePositionsArray
        let indexInPossiblePostionArray = possiblePositionsArray.indexOf(filledPositions[i])!

        if indexInPossiblePostionArray < possiblePositionsArray.count-2 && // safety check
            filledPositions[i+2].x < selectedElement.x &&  // Only check left of selected element
            //check equality of second items
            filledPositions[i+1].x == possiblePositionsArray[indexInPossiblePostionArray+1].x &&
            //check equality of third items
            filledPositions[i+2].x == possiblePositionsArray[indexInPossiblePostionArray+2].x {
            //3 consecutive elements to left selected element was found
            for j in i...i+2 {
                //add to left consecutive group
                consLeft.append(filledPositions[j])
            }
            //break out of the for loop
            break
        }
    }

    //The same thing in reversed order
    for i in (2..<filledPositions.count).reverse() {
        let indexInPossiblePostionArray = possiblePositionsArray.indexOf(filledPositions[i])!

        if indexInPossiblePostionArray-2 >= 0 &&
            filledPositions[i-2].x > selectedElement.x &&
            filledPositions[i-1].x == possiblePositionsArray[indexInPossiblePostionArray-1].x &&
            filledPositions[i-2].x == possiblePositionsArray[indexInPossiblePostionArray-2].x {
            for j in i-2...i {
                consRight.append(filledPositions[j])
            }
            break
        }
    }
}