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Swift 发布时间：2022-04-30 发布网站：大佬教程 code.js-code.com

大佬教程收集整理的这篇文章主要介绍了数组 – 如何从数组中找到最远的3个连续元素，大佬教程大佬觉得挺不错的，现在分享给大家，也给大家做个参考。

我有一组可能的位置和另一个填充位置数组,这是possiblePositionsArray的子数组. possiblePositionsArray是固定的,已经确定.我想在filledPositions中找到所选数组元素x位置右侧和左侧最远的3个连续点.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,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,则以下将是右侧和左侧的3个连续点.

[P7,p9] and [p8,p10] To the Left of p11
[p12,p14] and [p13,p15] to the right of p11

但是左边和右边最远的是：

farthest to left of p11 is [P7,p9]
farthest to right of p11 is [p13,p15]

我怎样才能做到这一点？

解决方法

首先从filledPositions开始.在possiblePositionsArray中找到fillPositions中的第一项.检查两个阵列中的下两个项是否相互匹配.第一个连续组是所选元素左侧最远的组.即使possiblePositionsArray元素中的x值不是等间距也是如此.

之后,你按相反的顺序执行此操作,以找到最右边的.

代码就是这样的：

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
        }
    }
}