标签：就是 list int div 根据 his span [] arrays
CopyOnWriteSet
看了下CopyOnWriteSet源码,底层使用的是CopyOnWriteList,根据底层的实现,每次读取都是N的复杂度.每次写也是N的复杂度.有个代码可以看一下,
来自CopyOnWriteList
public boolean addIfAbsent(E e) { Object[] snapshot = getArray(); return indexOf(e, snapshot, 0, snapshot.length) >= 0 ? false : addIfAbsent(e, snapshot); }
private boolean addIfAbsent(E e, Object[] snapshot) { final ReentrantLock lock = this.lock; lock.lock(); try { Object[] current = getArray(); int len = current.length; if (snapshot != current) { // Optimize for lost race to another addXXX operation int common = Math.min(snapshot.length, len); for (int i = 0; i < common; i++) if (current[i] != snapshot[i] && eq(e, current[i])) return false; if (indexOf(e, current, common, len) >= 0) return false; } Object[] newElements = Arrays.copyOf(current, len + 1); newElements[len] = e; setArray(newElements); return true; } finally { lock.unlock(); } }
代码的逻辑如下,获取数组snapshot,执行indexOf,如果没有,准备把新元素加入进去.
在实际加入方法内,获取锁,获取最新的数组current,如果俩个指针还是同一个指针,那就是没有改变,要不然.
取来个数组长度的最小值.
对current里面获取改变后的元素执行eq方法.对未比较过的元素执行indexof方法.要是还是没有重复的方法,就加入e
标签：就是 list int div 根据 his span [] arrays
原文地址：https://www.cnblogs.com/shuiyonglewodezzzzz/p/9446431.html
