c++ - 计算每个子串出现的次数？

Question

给定一个字符串 S，我想计算出现 n 次的子字符串的数量 (1 <= n <= s.length())。我已经用rolling hash完成了，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？

比如 s = "ababaab"

n 个字符串

4 1 "a"(子字符串 "a"出现了 4 次)

3 2 "b", "ab"(子字符串 "b"和 "ab"出现了 3 次)

2 2 "ba", "aba"

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

Answer 1

这不是一个获取免费代码的论坛，但由于我今晚的模组非常好，所以我为您写了一个简短的示例。但我不能保证没有错误，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;