多层划分&Bitmap


多层划分
1、2.5亿个整数中找出不重复的整数的个数,内存空间不足以容纳这2.5亿个整数
分析:有点像鸽巢原理,整数个数为2^32,也就是,我们可以将这2^32个数,划分为2^8个区域(比如用单个文件代表一个区域),然后将数据分离到不同的区域,然后不同的区域在利用bitmap就可以直接解决了。也就是说只要有足够的磁盘空间,就可以很方便的解决。
2、5亿个int找它们的中位数
分析:首先我们将int划分为2^16个区域,然后读取数据统计落到各个区域里的数的个数,之后我们根据统计结果就可以判断中位数落到那个区域,同时知道这个区域中的第几大数刚好是中位数。然后第二次扫描我们只统计落在这个区域中的那些数就可以了。
实际上,如果不是int是int64,我们可以经过3次这样的划分即可降低到可以接受的程度。即可以先将int64分成2^24个区域,然后确定区域的第几大数,在将该区域分成2^20个子区域,然后确定是子区域的第几大数,然后子区域里的数的个数只有2^20,就可以直接利用direct addr table进行统计了。

Bitmap
1、在2.5亿个整数中找出不重复的整数,注,内存不足以容纳这2.5亿个整数
http://m.blog.csdn.net/blog/chentao1206/40184419
解法一 :采用2-Bitmap(每个数分配2bit,00表示不存在,01表示出现一次,10表示多次,11无意义)进行,共需内存2^32 * 2 bit=1 GB内存,还可以接受。然后扫描这2.5亿个整数,查看Bitmap中相对应位,如果是00变01,01变10,10保持不变。所描完事后,查看bitmap,把对应位是01的整数输出即可。

解法二 :也可采用与第1题类似的方法,进行划分小文件的方法。然后在小文件中找出不重复的整数,并排序。然后再进行归并,注意去除重复的元素。”

2、给40亿个不重复的unsigned int的整数,没排过序的,然后再给一个数,如何快速判断这个数是否在那40亿个数当中?

解法一 :可以用位图/Bitmap的方法,申请512M的内存,一个bit位代表一个unsigned int值。读入40亿个数,设置相应的bit位,读入要查询的数,查看相应bit位是否为1,为1表示存在,为0表示不存在。
外排序
、已知某个文件内包含一些电话号码,每个号码为8位数字,统计不同号码的个数。 8位最多99 999 999,大概需要99m个bit,大概10几m字节的内存即可。

Bloom Filter
Bloom Filter,被译作称布隆过滤器,是一种空间效率很高的随机数据结构,Bloom filter可以看做是对bit-map的扩展,它的原理是:
  • 当一个元素被加入集合时,通过K个Hash函数将这个元素映射成一个位阵列(Bit array)中的K个点,把它们置为1**。检索时,我们只要看看这些点是不是都是1就(大约)知道集合中有没有它了:
    • 如果这些点有任何一个0,则被检索元素一定不在;
    • 如果都是1,则被检索元素很可能在。
其可以用来实现数据字典,进行数据的判重,或者集合求交集。
但Bloom Filter的这种高效是有一定代价的:在判断一个元素是否属于某个集合时,有可能会把不属于这个集合的元素误认为属于这个集合(false positive)。因此,Bloom Filter不适合那些“零错误”的应用场合。而在能容忍低错误率的应用场合下,Bloom Filter通过极少的错误换取了存储空间的极大节省。

1、给你A,B两个文件,各存放50亿条URL,每条URL占用64字节,内存限制是4G,让你找出A,B文件共同的URL。如果是三个乃至n个文件呢?

分析 :如果允许有一定的错误率,可以使用Bloom filter,4G内存大概可以表示340亿bit。将其中一个文件中的url使用Bloom filter映射为这340亿bit,然后挨个读取另外一个文件的url,检查是否与Bloom filter,如果是,那么该url应该是共同的url(注意会有一定的错误率)。”
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