下面是我使用的代码,on.it从字符串中获取单词计数,但现在我一直试图使用映射应用相同的逻辑,但无法这样做,因为在运行时,映射不能取键值,我是否可以每次从字符串中将每个单词存储在不同的键中,这样我就可以得到实际的单词计数,.Any,知道如何做到这一点吗?
#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] != '\0'; 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();
}发布于 2018-04-25 21:47:12
如何每次从不同键的字符串中将每个单词存储在键中,以便得到实际的单词计数。
这可以按以下方式进行。您甚至可以处理给定句子/字符串中的每个单词(假设每个单词被空格分隔)。
有几件事要注意:
#include<conio.h>已在您的soln中使用)using namespace std;std::map<>,则必须包括标题<map>例如,这里有一个示例测试输出: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;
}https://stackoverflow.com/questions/50027777
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