关于字符串的一个算法请教
假设我有这么一段字符串:
asda2121ffasas你好你好啊你好你好你好啊22111000csdnsdn
上面这个字符串处理后应该为:
asda2121ffasas你好你好啊你好你好啊221100csdnsdn
也就是说:
在这段字符串中,任何连续出现的字符或子字符串,不能大于两次,否则就只显示两次。
最差的算法应该是把字符串一层层便利,得出每一个字符及子字符串所出现的次数,有大于2的加以处理。但如果这个字符串很长的话,效率实在太低了。请教优化算法。谢谢。
其实这是我们网站的策划给我提的一个需求,因为有很多网友在留言的时候总是写一些重复的留言,他想让这些留言只显示一定次数多余的就不显示了。这个需求我已经给他打回去了,因为如果按照最差的算法效率实在太低,但还是想知道有没有什么比较高效的算法。
------解决方案--------------------
分词之后, 再去一个个遍历字符串, 基本上和你所说的很不效率, 如果放在服务器更是要命, 假使你真的想实现这个方法, 建议你把它放在客户端用JS实现。
当用户在编辑文本时, 监测到对方不判断ctrl+c和ctrl+v, 你就给它不停弹窗口,“哥们, csdn不可以灌水, 不可以回复内容太短噢 ... $&^#%&*(... ”
------解决方案--------------------
$s = 'asda2121ffasas你好你好啊你好你好你好啊22111000csdnsdn ';
echo preg_replace('/(.+)\\1+/', '$1$1', $s);
out:
asda2121ffasas你好你好啊你好你好啊221100csdnsdn
$s = 'asda2121ffasas你好你好啊好啊好啊你好你好你好啊22111000csdncsdncsdnsdn ';
echo preg_replace('/(.+)\\1+/', '$1$1', $s);
out:
asda2121ffasas你好你好啊好啊好啊你好你好啊221100csdncsdnsdn
------解决方案--------------------
这个算法没那么简单吧。O(n)估计没戏。
唠叨例子的第二次结果里出现了3次'好啊'了。
即使加入中文正则匹配
问题还有类似这样的
$str = '221112211122111';//'22111' 3次,'1' 3次
最终处理结果希望是什么样?22112211?
------解决方案--------------------
#23楼的思路回到while循环那去了,呵呵.
就如我#9想的一样.等同于
- PHP code
$s = '我是你你是我我是你你是我我是你你是我.我是你你是我我是你你是我我是你你是我.我是你你是我我是你你是我我是你你是我.我是你你是我我是你你是我我是你你是我'; $str = preg_replace('/(.+?)\\1{2,}/', '$1$1',$s); while(preg_match('/(.+?)\\1{2,}/', $str)) { $str = preg_replace('/(.+?)\\1{2,}/', '$1$1',$str); } echo $str; <div class="clear"> </div>

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