Let’s assume iterator B is currently positioned on 50, and iterator A is positioned on 100. In order to determine if 100 is in B, we have to advance B to 10000 . This can be an expensive operation because advancing to the next match may require scanning several blocks containing values between 50 and 10000.
11 - The Coherence Problem。关于这个话题，新收录的资料提供了深入分析
// This callback executes on game-loop thread.，推荐阅读新收录的资料获取更多信息
除了体验与价格，精细化、高共情的服务，更是新中式乐园留住游客的核心竞争力。