计算每个子串的出现次数?

时间:2015-06-11 17:51:25

标签: c++ string suffix-tree suffix-array

给定字符串 S ,我想计算 n 次(1< = n< = s.length())时出现的子串数。我用滚动哈希完成了它,它可以通过使用后缀树来完成。如何使用复杂度为O(n ^ 2)的后缀数组来解决?

喜欢s =" ababaab"

n no.of string

4 1" a" (子串" a"存在4次)

3 2" b" ," ab" (子串" b"和" ab"有3次出现)

2 2" ba" ," aba"

1 14" aa" ," bab" ," baa" ," aab" ," abab" ....

1 个答案:

答案 0 :(得分:1)

这不是一个获得免费代码的论坛,但由于我在这样一个很好的模式,我为你写了一个简短的例子。但我不能保证这是没有错误的,这是在15分钟内写的,没有特别多的想法。

#include <iostream>
#include <cstdlib>
#include <map>

class CountStrings
{
    private:
            const std::string               text;
            std::map <std::string, int>     occurrences;

            void addString ( std::string );
            void findString ( std::string );

    public:
            CountStrings ( std::string );
            std::map <std::string, int> count ( );
};

void CountStrings::addString ( std::string text)
{
    std::map <std::string, int>::iterator iter;

    iter = ( this -> occurrences ).end ( );

    ( this -> occurrences ).insert ( iter, std::pair <std::string, int> ( text, 1 ));
}

void CountStrings::findString ( std::string text )
{
    std::map <std::string, int>::iterator iter;

    if (( iter = ( this -> occurrences ).find ( text )) != ( this -> occurrences ).end ( ))
    {
            iter -> second ++;
    }
    else
    {
            this -> addString ( text );
    }
}

CountStrings::CountStrings ( std::string _text ) : text ( _text ) { }

std::map <std::string, int> CountStrings::count ( )
{
    for ( size_t offset = 0x00; offset < (( this -> text ).length ( )); offset ++ )
    {
            for ( size_t length = 0x01; length < (( this -> text ).length ( ) - (  offset - 0x01 )); length ++ )
            {
                    std::string subtext;

                    subtext = ( this -> text ).substr ( offset, length );

                    this -> findString ( subtext );
            }
    }

    return ( this -> occurrences );
}

int main ( int argc, char **argv )
{
    std::string text = "ababaab";
    CountStrings cs ( text );
    std::map <std::string, int> result = cs.count ( );

    for ( std::map <std::string, int>::iterator iter = result.begin ( ); iter != result.end ( ); ++ iter )
    {
            std::cout << iter -> second << " " << iter -> first << std::endl;
    }

    return EXIT_SUCCESS;

}