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c++ - 如何二进制搜索结构向量并在适当的索引处插入

问题描述

我想做的事：

我想要一个排序的文本表（大小相等），它显示每个文本的出现次数。在将文本插入表中时对表进行排序。当单词被插入到表中时，会检查它们是否已经存在，在这种情况下它们的ref_count会增加。如果它们是新的，则它们的ref_count设置为 1 并插入到正确索引处的向量中，以确保表仍然是排序的。

我做了什么：

我创建了一个结构体和一个结构体向量，定义如下。然后我使用显示的二进制搜索来识别适当的索引以使用std::insert()函数。

我的问题：我的二进制搜索实现没有返回正确的索引位置。

#define WORD_LENGTH 6

typedef struct RC_table {
    char word[WORD_LENGTH+1];//+1 for ‘\0’
    unsigned int ref_count;
}RC_table;

std::vector<RC_table>RC;

void update_RC(char *word_to_insert)
{
    int index = 0; 
    bool found=binary_search_RC(word_to_insert, &index);
    if (found == TRUE) {
        //increment reference count
        RC[index].ref_count++;
    }
    else {
        //insert new word at index 
        RC_table new_entry;
        memcpy(new_entry.word, word_to_insert, WORD_LENGTH);
        new_entry.word[WORD_LENGTH] = '\0';
        new_entry.ref_count = 1;

        if(index==0)
            RC.insert(RC.begin(),new_entry);
        else if(index==RC.size()-1)
            RC.insert(RC.end(),new_entry);
        else {
            RC.insert(RC.begin() + index +1, new_entry);
        }
    }
}
bool binary_search_RC(char *word, int *index) {
    int left = 0;
    int right = RC.size() -1;
    int middle = (left + right) / 2;
    bool found = false;
    while (left<=right) {
        middle = (left + right) / 2;
        *index = middle;
        if (strncmp(word, RC[middle].word, WORD_LENGTH) < 0) {
            right = middle - 1;
        }

        else if(strncmp(word, RC[middle].word, WORD_LENGTH) > 0) {
            left = middle + 1;
        }
        else {
            found = true;
            break;
        }
    }
    *index = middle;
    return found;
}

编辑： 我尝试使用lower_bound()。它仍然没有给出预期的输出（即排序表）。

typedef struct RC_table {
    char word[WORD_LENGTH+1];//+1 for ‘\0’
    unsigned int ref_count;
    bool operator<(const RC_table&r){
         return word<r.word;
    }
}RC_table;

使用以下方法插入表格：

auto itr=lower_bound(RC.begin(),RC.end(),new_enry);
RC.insert(itr,new_entry);

标签： c++vectorbinary-searchinsert-update

解决方案

您的算法的最大问题是您试图使用您认为可搜索的元素来隔离分区的两侧。这不是你这样做的方式；在最坏的情况下，分区可能（并且将会）最终陷入重复的 1/2 整数除法。当你这样做时，数学计算起来要容易得多：

  • 分区的左端指的是第一个元素，被认为是可疑的。
  • 分区的右端指的是过去的元素。并且不被视为可疑。

结果是一个更简单的算法，更容易理解，也更容易维护。

bool binary_search_RC(const char *word, int *index)
{
    *index = 0;

    int left = 0;
    int right = static_cast<int>(RC.size());
    bool found = false;
    while (!found && left < right)
    {
        int middle = *index = left + (right-left) / 2;
        int res = strncmp(word, RC[middle].word, WORD_LENGTH);

        if (res < 0)
            right = middle;

        else if (res > 0)
            *index = left = middle+1;

        else
            found = true;
    }
    return found;
}

付诸实践，一个简单的小测试工具，它从一个简单的三字母字母表中生成随机的三字符串。这应该会提供大量独特的插入，并最终会有大量的发现。最后，我们将打印整个表格，如果可行，最好对其进行排序。

代码

#include <iostream>
#include <vector>
#include <random>

#define WORD_LENGTH 6

typedef struct RC_table {
    char word[WORD_LENGTH+1];//+1 for ‘\0’
    unsigned int ref_count;
} RC_table;

std::vector<RC_table>RC;

bool binary_search_RC(const char *word, int *index)
{
    *index = 0;

    int left = 0;
    int right = static_cast<int>(RC.size());
    bool found = false;
    while (!found && left < right)
    {
        int middle = *index = left + (right-left) / 2;
        int res = strncmp(word, RC[middle].word, WORD_LENGTH);

        if (res < 0)
            right = middle;

        else if (res > 0)
            *index = left = middle+1;

        else
            found = true;
    }
    return found;
}

void update_RC(const char *word_to_insert)
{
    int index = 0;
    bool found = binary_search_RC(word_to_insert, &index);

    if (found)
    {
        ++RC[index].ref_count;
        std::cout << "Found entry for " << word_to_insert;
        std::cout << " (" << RC[index].ref_count << ")\n";
    }
    else {
        std::cout << "Adding entry for " << word_to_insert << '\n';

        //insert new word at index
        RC_table new_entry;
        strncpy(new_entry.word, word_to_insert, WORD_LENGTH);
        new_entry.word[WORD_LENGTH] = 0;
        new_entry.ref_count = 1;

        if(index==0)
            RC.insert(RC.begin(),new_entry);

        else if(index == RC.size())
            RC.insert(RC.end(),new_entry);

        else
            RC.insert(RC.begin() + index, new_entry);
    }
}



int main()
{
    // generate some random values and start adding them. a few dozen
    //  with a severely limited alphabet should suffice.
    const char alphabet[] = "abc";
    std::mt19937 rng{ std::random_device{}() };
    std::uniform_int_distribution<std::size_t> dist(0, sizeof alphabet - 2);

    for (int i=0; i<50; ++i)
    {
        char word[WORD_LENGTH+1] = {};
        for (int j=0; j<3; ++j)
            word[j] = alphabet[ dist(rng) ];
        update_RC(word);
    }

    // print the table
    for (auto const& x : RC)
        std::cout << x.word << " : " << x.ref_count << '\n';
}

输出（显然不同）

Adding entry for cab
Adding entry for cac
Adding entry for bcc
Adding entry for bbb
Adding entry for cbc
Adding entry for abb
Found entry for cab (2)
Adding entry for aba
Adding entry for cca
Adding entry for acc
Found entry for aba (2)
Found entry for bcc (2)
Adding entry for cbb
Found entry for cac (2)
Found entry for cac (3)
Adding entry for aaa
Found entry for acc (2)
Adding entry for bbc
Adding entry for baa
Adding entry for acb
Found entry for aaa (2)
Found entry for cca (2)
Found entry for baa (2)
Found entry for cbb (2)
Adding entry for aac
Found entry for cac (4)
Adding entry for aca
Adding entry for ccc
Found entry for bbc (2)
Adding entry for bba
Adding entry for bac
Adding entry for aab
Found entry for bac (2)
Found entry for aca (2)
Found entry for bcc (3)
Adding entry for caa
Found entry for aaa (3)
Found entry for bbc (3)
Found entry for caa (2)
Found entry for abb (2)
Found entry for baa (3)
Found entry for acc (3)
Found entry for bba (2)
Found entry for bbb (2)
Found entry for cbc (2)
Found entry for aaa (4)
Found entry for baa (4)
Adding entry for cba
Found entry for bac (3)
Found entry for bbc (4)
aaa : 4
aab : 1
aac : 1
aba : 2
abb : 2
aca : 2
acb : 1
acc : 3
baa : 4
bac : 3
bba : 2
bbb : 2
bbc : 4
bcc : 3
caa : 2
cab : 2
cac : 4
cba : 1
cbb : 2
cbc : 2
cca : 2
ccc : 1

我没有费心做数学，但是把这些引用计数加起来，你应该会发现它们总和为 50，即我们执行的插入次数。

祝你好运。

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