我在
JavaScript中实现了以下链表数据结构:
class Node {
constructor(data, list) {
this.data = data;
this.list = list;
this.prev = null;
this.next = null;
}
remove() {
if (this.prev) {
this.prev.next = this.next;
} else {
this.list.start = this.next;
}
if (this.next) {
this.next.prev = this.prev;
} else {
this.list.end = this.prev;
}
this.next = null;
this.prev = null;
this.list.length -= 1;
}
}
class LinkedList {
constructor() {
this.end = null;
this.start = null;
this.length = 0;
}
append(data) {
const node = new Node(data, this);
if (!this.start) {
this.start = node;
}
if (this.end) {
node.prev = this.end;
this.end.next = node;
}
this.end = node;
this.length += 1;
return data;
}
remove() {
if (this.end) {
return this.end.remove();
}
}
*[Symbol.iterator]() {
let current = this.start;
while (current) {
yield current;
current = current.next;
}
}
}
module.exports = LinkedList;
我这样使用它来更新动画列表:
static update(timeDelta) {
for (let node of this.animations) {
const animation = node.data;
if (animation.animating) {
animation.update(timeDelta);
} else {
node.remove();
}
}
}
node.remove()行在我的游戏中引起非常明显的延迟.我怀疑它正在触发垃圾收集.反过来说,如果我注释掉node.remove()行并允许链表永远增长,游戏就会顺利运行.
动画不断被添加和删除.我在动画更新功能中添加了一些日志记录:
start iterating linked list
removing
ms elapsed: 0.45499999999992724
end iterating
start iterating linked list
removing
ms elapsed: 0.455000000000382
end iterating
start iterating linked list
removing
ms elapsed: 0.13000000000010914
end iterating
start iterating linked list
(13) updating
ms elapsed: 2.200000000000273
end iterating
您可以看到链接列表每秒迭代多次,偶尔会删除节点.
如何在不实际导致性能下降的情况下从列表中删除O(1)?
最佳答案
The
node.remove()
line causes very noticeable lag in my game. I suspect it is triggering garbage collection.
不.滞后来自每个更新调用仅更新极少数动画的事实.
您的问题是旧的和着名的“迭代期间删除”问题.在您的情况下,它不会触发罕见的边缘情况错误,它只是停止迭代:
while (current) {
yield current;
// after yielding, in the `update` function, we have
// node = current
// and most importantly
// node.remove()
// which assigns (with `this` being `node`)
// this.next = null;
// then the iteration is resumed
current = current.next;
}
哎呀.简单的解决方法是在产生之前缓存要迭代的下一个节点:
let next = this.start;
while (next) {
const current = next;
next = current.next;
yield current;
}
(或类似的东西),但当然删除下一个节点时仍然会失败.更好的方法可能是省略线条
this.next = null;
this.prev = null;
来自节点的remove方法,以便在删除期间引用保持不变.这不会影响GC.
另一种解决方案是完全删除链接列表 – 除非您经常在迭代之外添加/删除列表中间的节点,否则它会过度设计.在迭代过程中过滤旧的动画很简单,它可以使用一个好的旧(内存效率高)?数组来完成,甚至是就地:
function filter(array, callback) {
var i=0, j=0;
while (j < array.length) {
if (callback(array[j]))
array[i++] = array[j++];
else
array[i] = array[j++];
}
array.length = i;
}
function update(timeDelta) {
filter(animations, animation => {
var keep = animation.animating;
if (keep) animation.update(timeDelta);
return keep;
});
}
(当你= = j时,你可以通过不重新分配来优化过滤器)