我的目的是在曲线的中点绘制一个物体.使用SVG DOM,我可以使用方法getPointAtLength& getTotalLength,但我看不到HTML画布中的等价物.

HTML画布中没有等效的东西.你必须用简单的旧数学自己找到中点.

我做了一个如何为你找到贝塞尔曲线中点的例子.在jsfiddle here上看到它.下面粘贴了一个javascript的副本.

实曲线为红色,中点为微小的绿色矩形.其他一切只是一种视觉辅助.

var ctx = $("#test")[0].getContext("2d")
function mid(a,b) {
 return (a+b) / 2;
}


var cp1x = 100;
var cp1y = 150;
var cp2x = 175;
var cp2y = 175;
var x = 200;
var y = 0;

ctx.linewidth = 4;
ctx.stroke;
ctx.fill;

ctx.beginPath();
ctx.moveto(0,0);
ctx.bezierCurveto(cp1x,cp1y,cp2x,cp2y,x,y);
ctx.stroke();

//line goes from start to control point 1
ctx.stroke;
ctx.beginPath();
ctx.moveto(0,0);
ctx.lineto(cp1x,cp1y);
ctx.stroke();

//line goes from end to control point 2
ctx.beginPath();
ctx.moveto(x,y);
ctx.lineto(cp2x,cp2y);
ctx.stroke();

//line goes from control point to control point
ctx.stroke;
ctx.beginPath();
ctx.moveto(cp1x,cp1y);
ctx.lineto(cp2x,cp2y);
ctx.stroke();

// Now find the midpoint of each of those 3 lines
var ax = mid(cp1x,0);
var bx = mid(cp2x,x)
var cx = mid(cp1x,cp2x)

var ay = mid(cp1y,0)    
var by = mid(cp2y,y)    
var cy = mid(cp1y,cp2y)


// draw midpoints for visual aid 
// not gonna look exact 'cause square
// will be drawn from top-right instead of center
ctx.fillRect(ax,ay,4,4);
ctx.fillRect(bx,by,4);
ctx.fillRect(cx,cy,4);


//Now draw lines between those three points. These are green
ctx.stroke;
ctx.beginPath();
ctx.moveto(ax,ay);
ctx.lineto(cx,cy);
ctx.stroke();

ctx.beginPath();
ctx.moveto(bx,by);
ctx.lineto(cx,cy);
ctx.stroke();

//Now the midpoint of the green lines:
// so g1 and g2 are the green line midpoints
var g1x = mid(ax,cx);
var g2x = mid(bx,cx);

var g1y = mid(ay,cy); 
var g2y = mid(by,cy);  

//draw them to make sure:
ctx.fillRect(g1x,g1y,4);
ctx.fillRect(g2x,g2y,4);

//Now one final line,in gray
ctx.stroke;
ctx.beginPath();
ctx.moveto(g1x,g1y);
ctx.lineto(g2x,g2y);
ctx.stroke();

//whew! We made it!
var FinallyTheMidpointx = mid(g1x,g2x); 
var FinallyTheMidpointy = mid(g1y,g2y); 

//draw something at the midpoint to celebrate
ctx.fill;
ctx.fillRect(FinallyTheMidpointx,FinallyTheMidpointy,4);

以下所有产生可接受的结果. u1似乎比预期的列表大小(约1000-2000 KVP)最快

我们能做得更好(更快)吗？

private static Set<String> u1(List<_KVPair> pairs) {
    Set<String> undefined = new HashSet<String>();

    for (_KVPair pair : pairs) {
        undefined.add(pair.getValue());
    }

    if (undefined.size() == 1) {
        return new HashSet<String>();
    }
    return undefined;
}

private static List<String> u2(List<_KVPair> pairs) {

    List<String> undefined = new ArrayList<String>();
    for (_KVPair pair : pairs) {
        if (!undefined.contains(pair.getValue())) {
            undefined.add(pair.getValue());
        }
    }

    return undefined;
}

private static List<String> u3(List<_KVPair> pairs) {

    List<String> undefined = new LinkedList<String>();

    Iterator<_KVPair> it = pairs.iterator();
    while (it.hasNext()) {
        String value = it.next().getValue();
        if (!undefined.contains(value)) {
            undefined.add(value);
        }
    }
    return undefined;
}

大约3600对,’u3’获胜.大约1500对,’u1’获胜

Set<String> undefined = new HashSet<String>(pairs.size(),1);

请注意,我使用1作为加载因子来防止任何大小调整.

出于好奇,我进行了测试(下面的代码) – 结果是(编译后)：

测试1(注意：热身需要几分钟)

size of original list = 3,000 with no duplicates:
set: 8
arraylist: 668
linkedlist: 1166

测试2

size of original list = 30,000 – all strings identical:
set: 25
arraylist: 11
linkelist: 13

那种有道理：

>当有许多重复项时,List#contains将运行得相当快,因为​​可以更快地找到重复项并且分配大型集合哈希算法的成本正在受到影响
>当没有或很少重复时,该集合大幅度地获胜.

public class TestPerf {

    private static int NUM_RUN;
    private static Random r = new Random(System.currentTimeMillis());
    private static boolean random = false; //toggle to false for no duplicates in original list


    public static void main(String[] args) {

        List<String> list = new ArrayList<>();

        for (int i = 0; i < 30_000; i++) {
            list.add(getRandomString());
        }

        //warm up
        for (int i = 0; i < 10_000; i++) {
            method1(list);
            method2(list);
            method3(list);
        }

        NUM_RUN = 100;
        long sum = 0;
        long start = System.nanoTime();
        for (int i = 0; i < NUM_RUN; i++) {
            sum += method1(list);
        }
        long end = System.nanoTime();
        System.out.println("set: " + (end - start) / 1000000);

        sum = 0;
        start = System.nanoTime();
        for (int i = 0; i < NUM_RUN; i++) {
            sum += method2(list);
        }
        end = System.nanoTime();
        System.out.println("arraylist: " + (end - start) / 1000000);

        sum = 0;
        start = System.nanoTime();
        for (int i = 0; i < NUM_RUN; i++) {
            sum += method3(list);
        }
        end = System.nanoTime();
        System.out.println("linkelist: " + (end - start) / 1000000);

        System.out.println(sum);
    }

    private static int method1(final List<String> list) {
        Set<String> set = new HashSet<>(list.size(),1);
        for (String s : list) {
            set.add(s);
        }
        return set.size();
    }

    private static int method2(final List<String> list) {
        List<String> undefined = new ArrayList<>();
        for (String s : list) {
            if (!undefined.contains(s)) {
                undefined.add(s);
            }
        }
        return undefined.size();
    }

    private static int method3(final List<String> list) {
        List<String> undefined = new LinkedList<>();

        Iterator<String> it = list.iterator();
        while (it.hasNext()) {
            String value = it.next();
            if (!undefined.contains(value)) {
                undefined.add(value);
            }
        }
        return undefined.size();
    }

    private static String getRandomString() {
        if (!random) {
            return "skdjhflkjrglajhsdkhkjqwhkdjahkshd";
        }
        int size = r.nextInt(100);
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < size; i++) {
            char c = (char) ('a' + r.nextInt(27));
            sb.append(c);
        }
        System.out.println(sb);
        return sb.toString();
    }
}