fiz*_*zis 11 javascript recursion
我有一个递归功能,可以在画布上移动一些圆圈.被遮挡的圆圈被放大(放大),所有其他圆圈被推开.推动圆圈推动其他圆圈,依此类推,直到缩放完成.
我得到一个错误"超出最大调用堆栈大小",我理解这个问题,但我只是不知道如何解决它...我找到了三种解决递归问题的可能解决方案:
但我认为我不能使用它们:
我该如何解决这个问题?
// Pushes circles aside when some other circle leans on these circles (on zoom in)
var moveCirclesAside = function(circle1, circleToSkip, groupOfMoves) {
var count = circles.length;
for (var i = 0; i < count; i++) {
// Skip the same circle
if (i == circle1.i) {
continue;
}
// Also skip the circle which was intended not to move any further
if (circleToSkip != null && i == circleToSkip.i) {
continue;
}
// Get second circle
var circle2 = circles[i];
// Calculate a distance between two circles
var dx = circle2.x - circle1.x;
var dy = circle2.y - circle1.y;
var distance = Math.sqrt((dx * dx) + (dy * dy));
// If circles already collided need to do some moving...
if (distance <= circle1.r + circle2.r + OD.config.circleSpacing) {
// Get collision angles
var angle = Math.atan2(dy, dx);
var sine = Math.sin(angle);
var cosine = Math.cos(angle);
// Some circle position calculation
var x = OD.config.circleSpacing;
var xb = x + (circle1.r + circle2.r);
var yb = dy * cosine - dx * sine;
// Save each state (move) of any circle to the stack for later rollback of the movement
groupOfMoves.push(copyCircleByVal(circle2));
// Move the circle
circle2.x = circle1.x + (xb * cosine - yb * sine);
circle2.y = circle1.y + (yb * cosine + xb * sine);
// Make sure that circle won't go anywhere out of the canvas
adjustCircleByBoundary(circle2);
// If moved circle leans against some other circles make sure that they are moved accordingly
// And such related moves must be grouped for correct rolback of moves later - so we pass 'groupOfMoves' var
moveCirclesAside(circle2, circle1, groupOfMoves);
}
}
};
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1)由于需要的操作数量未知,我无法实现迭代;
好吧,我没有查看你的代码,但一般避免线性递归(这里你有一个线性的)看起来像这样:
while (1 == 1) {
if (breakcondition)
break;
doSomeCode()
}
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因此,您不必知道for- 循环情况下的确切迭代次数.
这种溢出并不令人感到意外,因为算法在迭代时会增加堆栈,但退出条件是不可预测的,动作没有紧密定位(它们对附近的圆圈具有连锁效应),因此处理时间将是混乱的.
我会重新考虑这个算法.考虑找到两个最接近的圆圈,如果它们比分开的给定阈值更远,则中止.否则将它们分开并重复一遍.
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