寻找地图技术
looking for maps techniques
这是我正在使用的代码 on.it 从字符串中获取字数,但现在我一直在尝试使用地图应用相同的逻辑,但无法这样做,因为地图无法获取键值在 运行 time.how 我可以每次将字符串中的每个单词存储在不同键中的键中,以便我可以获得实际的字数。知道我该怎么做吗?
#include<iostream>
#include<conio.h>
#include<string>
using namespace std;
int main()
{
map<string, int> stringCounts;
map<string, int>::iterator iter;
string words;
int TOTAL = 0;
char a[1000];
cout << "enter the string = ";
cin.getline(a, 1000);
int Totalwords = 0;
int no = 0;
for (int i = 0; a[i] != '[=10=]'; i++)
{
if ((int(a[i]) >= 65 && int(a[i]) <= 90) || (int(a[i]) >= 97 && int(a[i]) <= 122))
{
}
else
{
Totalwords++;
}
no = i;
}
TOTAL = Totalwords;
cout << "number of words = " << TOTAL << endl;
string *words = new string[TOTAL];
for (int i = 0, j = 0; j < TOTAL, i <= no;)
{
if ((int(a[i]) >= 65 && int(a[i]) <= 90) || (int(a[i]) >= 97 && int(a[i]) <= 122))
{
words[j] = words[j] + a[i];
stringCounts[words[j]]++;
for (iter = stringCounts.begin(); iter != stringCounts.end(); iter++)
{
cout << "word: " << iter->first << ", count: " << iter->second <<
endl;
}
i++;
}
else
{
j++;
i++;
}
}
_getch();
}
how can i store each word in a key from a string in a different keys
each time so that i can get actual word count .
这可以按如下方式完成。您甚至可以处理给定句子/字符串的每个单词(假设每个单词都被 space 分隔)。
注意几点:
- 如果您使用 C++,请坚持使用 C++ std 库(
#include<conio.h>
已在您的解决方案中使用)
- 避免练习
using namespace std;
- 如果你想使用
std::map<>
,你必须包含 header <map>
例如,这是一个示例测试输出:https://www.ideone.com/KGua1M
#include <iostream>
#include <map>
#include <string>
#include <sstream>
int main()
{
std::string inputString;
std::cout << "Enter the string = ";
std::getline(std::cin, inputString);
std::map<std::string, int> Map; // word, no. of times
size_t wordCount = 0;
size_t letterCount = 0;
std::stringstream sstr(inputString);
std::string word;
while (std::getline(sstr, word, ' '))
{
Map[word]++;
wordCount++;
letterCount += word.size();
}
std::cout << "Total Words: " << wordCount << "\n\n";
std::cout << "Total letters: " << letterCount << "\n\n";
std::cout << "Each words count\n\n" ;
for(const auto& it: Map)
std::cout << it.first << " " << it.second << " times\n";
return 0;
}
这是我正在使用的代码 on.it 从字符串中获取字数,但现在我一直在尝试使用地图应用相同的逻辑,但无法这样做,因为地图无法获取键值在 运行 time.how 我可以每次将字符串中的每个单词存储在不同键中的键中,以便我可以获得实际的字数。知道我该怎么做吗?
#include<iostream>
#include<conio.h>
#include<string>
using namespace std;
int main()
{
map<string, int> stringCounts;
map<string, int>::iterator iter;
string words;
int TOTAL = 0;
char a[1000];
cout << "enter the string = ";
cin.getline(a, 1000);
int Totalwords = 0;
int no = 0;
for (int i = 0; a[i] != '[=10=]'; i++)
{
if ((int(a[i]) >= 65 && int(a[i]) <= 90) || (int(a[i]) >= 97 && int(a[i]) <= 122))
{
}
else
{
Totalwords++;
}
no = i;
}
TOTAL = Totalwords;
cout << "number of words = " << TOTAL << endl;
string *words = new string[TOTAL];
for (int i = 0, j = 0; j < TOTAL, i <= no;)
{
if ((int(a[i]) >= 65 && int(a[i]) <= 90) || (int(a[i]) >= 97 && int(a[i]) <= 122))
{
words[j] = words[j] + a[i];
stringCounts[words[j]]++;
for (iter = stringCounts.begin(); iter != stringCounts.end(); iter++)
{
cout << "word: " << iter->first << ", count: " << iter->second <<
endl;
}
i++;
}
else
{
j++;
i++;
}
}
_getch();
}
how can i store each word in a key from a string in a different keys each time so that i can get actual word count .
这可以按如下方式完成。您甚至可以处理给定句子/字符串的每个单词(假设每个单词都被 space 分隔)。
注意几点:
- 如果您使用 C++,请坚持使用 C++ std 库(
#include<conio.h>
已在您的解决方案中使用) - 避免练习
using namespace std;
- 如果你想使用
std::map<>
,你必须包含 header<map>
例如,这是一个示例测试输出:https://www.ideone.com/KGua1M
#include <iostream>
#include <map>
#include <string>
#include <sstream>
int main()
{
std::string inputString;
std::cout << "Enter the string = ";
std::getline(std::cin, inputString);
std::map<std::string, int> Map; // word, no. of times
size_t wordCount = 0;
size_t letterCount = 0;
std::stringstream sstr(inputString);
std::string word;
while (std::getline(sstr, word, ' '))
{
Map[word]++;
wordCount++;
letterCount += word.size();
}
std::cout << "Total Words: " << wordCount << "\n\n";
std::cout << "Total letters: " << letterCount << "\n\n";
std::cout << "Each words count\n\n" ;
for(const auto& it: Map)
std::cout << it.first << " " << it.second << " times\n";
return 0;
}