这篇文章主要介绍多段线实现布尔运算的方法
先上代码

function getOperatedCurves(sourceCurs: Curve[], targetCus: Curve[]){let source: Polyline | Circle = (sourceCurs[0] instanceof Circle) ? sourceCurs[0] as Circle : new Polyline().Combine(sourceCurs)[0];let target: Polyline | Circle = (targetCus[0] instanceof Circle) ? targetCus[0] as Circle : new Polyline().Combine(targetCus)[0];try{if (!source.IsClose || !target.IsClose) throw new Error("不是封闭曲线");}catch (err){console.log(err);}let interPts = source.IntersectWith(target, IntersectOption.OnBothOperands);let sourceContainerTarget = isTargetCurInSourceCur(source, target);let targetContainerSource = isTargetCurInSourceCur(target, source);let isContainer = sourceContainerTarget || targetContainerSource;let intersectionList: Curve[] = [];  //交集let unionList: Curve[] = []; //并集let subList: Curve[] = []; //补集/**两封闭区域有交点并且不是包含关系,则通过交点把区域分割*/if (interPts.length && !isContainer){let pars1 = interPts.map(p => source.GetParamAtPoint(p)).sort((a, b) => a - b);let pars2 = interPts.map(p => target.GetParamAtPoint(p)).sort((a, b) => a - b);let cus1: Array<Polyline | Arc> = source.GetSplitCurves(pars1);cus1.forEach(pl =>{if (isTargetCurInSourceCur(target, pl)){intersectionList.push(pl);}else{subList.push(pl);unionList.push(pl);}})let cus2: Array<Polyline | Arc> = target.GetSplitCurves(pars2);cus2.forEach(pl =>{if (isTargetCurInSourceCur(source, pl)){intersectionList.push(pl);subList.push(pl);}else{unionList.push(pl);}})}else{if (isContainer){if (sourceContainerTarget){intersectionList.push(target);subList.push(source, target);unionList.push(source);}else{unionList.push(target);intersectionList.push(source);}}else{unionList.push(source, target)subList.push(source);}}return {intersectionList, unionList, subList}}

由于一些曲线类实现方法不一，这里主要说一些实现布尔运算的思路

• 判断2封闭曲线是否是被包含的关系
• 获取2封闭曲线的所有交点，这里交点可能是圆和线，线和线，圆和圆的，求交点的方法网上应该很多，以后有时间也会写写用JavaScript实现方式
• 根据所有的交点把2封闭曲线分割为多个部分
• 对分割后的线段进行整理，其中相交部分是曲线在对方曲线内部的部分，合并是互不在对方曲线内部的部分，相减类似不想说了，具体看代码，如果是被包含状态则更加就简单了