tag:blogger.com,1999:blog-2084836737143870236.post6686931132351400152..comments2023-07-31T02:13:31.085-07:00Comments on 喜刷刷: [LeetCode新题] Read N Characters Given Read4 Yanbing Shihttp://www.blogger.com/profile/16541941455216571610noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-2084836737143870236.post-12164955226378145822017-12-07T00:25:48.378-08:002017-12-07T00:25:48.378-08:00请问怎么分析时间复杂度呢? 请问怎么分析时间复杂度呢? Anonymoushttps://www.blogger.com/profile/00012389960310819033noreply@blogger.comtag:blogger.com,1999:blog-2084836737143870236.post-51264407387230913572016-09-30T10:43:20.641-07:002016-09-30T10:43:20.641-07:00I think 4 is OKI think 4 is OK伍林森https://www.blogger.com/profile/17872086472911015896noreply@blogger.comtag:blogger.com,1999:blog-2084836737143870236.post-38193193805323133522014-12-16T00:59:31.150-08:002014-12-16T00:59:31.150-08:00HI May I ask two questions?
char *remain = new c...HI May I ask two questions?<br /><br /> char *remain = new char[5]; // why 5?<br /> m = min(read4(remain),n-len); // why are you doing this?寒雨飘零https://www.blogger.com/profile/18367780927983907852noreply@blogger.com