Question

有人可以回答以下C ++面试问题：

鉴于字符串：＆＃34;歌曲在印地文唱歌＆＃34;

找到重复的字符串，如下所示：

single characters repetition:

＆＃34; s＆＃34; = 2＆＃34; o＆＃34; = 0＆＃34; n＆＃34; = 5＆＃34; g＆＃34; = 3 .....等等。

two character repetition

＆＃34;所以＆＃34; = 0＆＃34; on＆＃34; = 0＆＃34; ng＆＃34; = 3＆＃34; in＆＃34; = 4 .....等

Three character repetition

＆＃34;儿子＆＃34; = 0 ....＆＃34; ing＆＃34; = 1 ....等

Four character repetition

＆＃34;歌曲＆＃34; = 0 .....等

RS

Answer 1

这是一种直截了当的方法

#include <iostream>
#include <string>
#include <map>

int main() 
{
    std::string s =  "song singing in hindi";

    for ( std::string::size_type i = 1; i <= s.size(); i++ )
    {
        std::map<std::string, size_t> m;

        for ( std::string::size_type j = 0; j < s.size() - i + 1; j++ )
        {
            m[std::string( s, j, i )]++;
        }

        for ( const auto &p : m )
        {
            std::cout << "( \"" << p.first << "\", " << p.second << " ) ";
        }
        std::cout << std::endl;
    }

    return 0;
}

如果要排除带有嵌入式空白的模式，可以按以下方式重写程序

#include <iostream>
#include <string>
#include <map>

int main() 
{
    std::string s =  "song singing in hindi";

    for ( std::string::size_type i = 1; i <= s.size(); i++ )
    {
        std::map<std::string, size_t> m;

        for ( std::string::size_type j = 0; j < s.size() - i + 1; j++ )
        {
            std::string t( s, j, i );
            if ( t.find( ' ' ) == std::string::npos )
            {
                m[t]++;
            }
        }

        if ( !m.empty() )
        {
            for ( const auto &p : m )
            {
                std::cout << "( \"" << p.first << "\", " << p.second << " ) ";
            }
            std::cout << std::endl;
        }
    }

    return 0;
}

输出

( "d", 1 ) ( "g", 3 ) ( "h", 1 ) ( "i", 5 ) ( "n", 5 ) ( "o", 1 ) ( "s", 2 ) 
( "di", 1 ) ( "gi", 1 ) ( "hi", 1 ) ( "in", 4 ) ( "nd", 1 ) ( "ng", 3 ) ( "on", 1 ) ( "si", 1 ) ( "so", 1 ) 
( "gin", 1 ) ( "hin", 1 ) ( "ind", 1 ) ( "ing", 2 ) ( "ndi", 1 ) ( "ngi", 1 ) ( "ong", 1 ) ( "sin", 1 ) ( "son", 1 ) 
( "ging", 1 ) ( "hind", 1 ) ( "indi", 1 ) ( "ingi", 1 ) ( "ngin", 1 ) ( "sing", 1 ) ( "song", 1 ) 
( "hindi", 1 ) ( "ingin", 1 ) ( "nging", 1 ) ( "singi", 1 ) 
( "inging", 1 ) ( "singin", 1 ) 
( "singing", 1 )

Answer 2

甚至递归函数都可以。删除＆＃39;模式＆＃39;使用空白区域，您可以使用字符串:: find函数来跟随Vlad的方法。

#include <iostream>
#include <string>
#include <map>

class FRQ {
    void Func(std::map<std::string, size_t> &frequencyTable, const std::string &m_str, const int stepping = 1) {
        if (stepping == m_str.size()) {
            frequencyTable[m_str]++;
            return;
        }

        for (std::string::size_type i = 0, iMAX = m_str.size(); i < iMAX; ++i) {
            frequencyTable[m_str.substr(i, stepping)]++;
        }

        Func(frequencyTable, m_str, stepping + 1);
    }
public:
    std::map<std::string, size_t> *operator()(const std::string &str) {
        std::map<std::string, size_t> *fTable = new std::map<std::string, size_t>();

        Func(*fTable, str);

        return fTable;
    }
};

int main(void) {
    using namespace std;

    string s = "HiYo HiYo";

    FRQ frq;
    map<string, size_t> *frequenceTable = frq(s);

    cout << "Patterns: " << frequenceTable->size() << endl;
    for (const auto& ptr : *frequenceTable)
        cout << "[ '" << ptr.first << "'::" << ptr.second << " ]" << endl;

    delete frequenceTable;

    return 0;
}

查找给定字符串中字符串模式的重复

2 个答案: