首页| JavaScript| HTML/CSS| Matlab| PHP| Python| Java| C/C++/VC++| C#| ASP| 其他|
购买积分 购买会员 激活码充值

您现在的位置是:虫虫源码 > C/C++/VC++ > 在一个无限长的字符流发展模式匹配

在一个无限长的字符流发展模式匹配

资 源 简 介

Programming Logic Test Using any common programming language, program a quick and straightforward answer to the following question (e.g. C++). A regular expression or other pattern matching package should not be used. Presented with a character stream that is infinite, output a "D" if the sequence "ddd" is located, precisely. Alternatively, if the characters "ded" are located, output an "E". You should not output both a "D" and an "E" for the same data when processed. For example: 1. The following stream of characters ddededddeddd would output: EDD 2. The following stream of characters dddededdddeedede would output: DEDE @author Gaurav Pandey Programming Logic: Queue (FIFO) is used to manage the sliding window which is of length 3 and wo

文 件 列 表

README.txt
LogicTest.jar
Source File
LogicTest.java
dddded_list.txt
Sample_Steps.txt

相 关 资 源

您 可 能 感 兴 趣 的

同 类 别 推 荐

VIP VIP