计算出线在屏幕内的最长坐标

原创于 2018-12-03 15:33:00 发布 · 150 阅读

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本文介绍了一种计算线段在屏幕内的最长显示部分的方法，通过解析线段两端坐标，利用二元一次方程原理，确定线段与屏幕边界相交的位置，从而找到线段在屏幕内的最大可见部分。

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计算出线在屏幕内的最长坐标

/**
* 计算出线在屏幕内的最长坐标
* x1, y1 = 坐标1, x2, y2 = 坐标2
*/
private createAimLinePosArr(x1: number, y1: number, x2: number, y2: number) {
    let posX, posY;
    if (y2 == y1) y1 += 0.01
    if (x2 == x1) x1 += 0.01

    if (y2 < y1) {
        if (x2 < x1) {
            posX = 0
            posY = this.binaryEquationGetY(x1, y1, x2, y2, posX)
        } else {
            posX = Const.STAGE_WIDTH //最大屏幕宽度
            posY = this.binaryEquationGetY(x1, y1, x2, y2, posX)
        }
    } else if (y2 > y1) {
        if (x2 > x1) {
            posX = Const.STAGE_WIDTH //最大屏幕宽度
            posY = this.binaryEquationGetY(x1, y1, x2, y2, posX)
        } else {
            posX = 0
            posY = this.binaryEquationGetY(x1, y1, x2, y2, posX)
        }
    }

    if (posY < Const.STAGE_HEIGHT) {
        if (posY <= 0) {
            posY = 0
            posX = this.binaryEquationGetX(x1, y1, x2, y2, posY)
        }
        return [posX, posY]
    } else {
        posY = Const.STAGE_HEIGHT //最大屏幕高度
        posX = this.binaryEquationGetX(x1, y1, x2, y2, posY)
        if (isNaN(posX)) {
            if (x2 > x1) {
                posX = Const.STAGE_WIDTH
            } else {
                posX = 0
            }
            console.log("posX == NaN", x1, y1, x2, y2, posY)
        }
        return [posX, posY]
    }
}

/**
 * 二元一次方程：根据x求y
 */
public binaryEquationGetY(x1, y1, x2, y2, x) {
    let kbArr = this.binaryEquationGetKB(x1, y1, x2, y2)
    let k = kbArr[0]
    let b = kbArr[1]
    return k * x + b
}

/**
 * 二元一次方程：根据y求x
 */
public binaryEquationGetX(x1, y1, x2, y2, y) {
    let kbArr = this.binaryEquationGetKB(x1, y1, x2, y2)
    let k = kbArr[0]
    let b = kbArr[1]
    return (y - b) / k
}

/**
 * 求二元一次方程的系数
 * y1 = k * x1 + b => k = (y1 - b) / x1
 * y2 = k * x2 + b => y2 = ((y1 - b) / x1) * x2 + b
 */
private binaryEquationGetKB(x1, y1, x2, y2) {
    let k = (y1 - y2) / (x1 - x2)
    let b = (x1 * y2 - x2 * y1) / (x1 - x2)
    return [k, b]
}