摘要: http://www.blogjava.net/lcs/archive/2007/10/09/151252.html 阅读全文
posted @ 2014-02-25 16:15 keia 阅读(61) 评论(0) 推荐(0) 编辑
摘要: long start = System.currentTimeMillis(); for (int i = 0; i < 1000000; i++) { } long end = System.currentTimeMillis(); System.out.println("---:"+(end-start)); long start1 = System.currentTimeMillis(); for (int i = 0; i < 1000000; i++) { try { } catch (Exception e) { } } long end1 = Sy 阅读全文
posted @ 2014-02-25 12:42 keia 阅读(379) 评论(0) 推荐(0) 编辑