C ++中的字符计数器？

Question

I'm trying to find a way to display all the characters in a string and the number of times they occur.

This is what I have so far:

//Any unused includes are part of the default code

#include <iostream>
#include <ctime>
#include <cstdlib>
#include <cmath>
#include <string>

using namespace std;

int main()
{
    string st = "";

    cout << "Input a sentence: " << endl;
    getline(cin, st);
    int index = 0;
    int index2 = 0;
    int counters[26] = {0};
    for(int i = 0; i < st.length(); i++)
    {
        int counter = 0;
        index = st.find(st[i],0);
        for(int j = 0; j < st.length(); j++)
        {
            index2 = st.find(st[j]);
            if(index == index2)
            {
                counter++;
            }
        }
        cout << st[i] << ": " << counter << endl;
    }
    //cout << st[i] <<": " << counters[st[i] - 'a'] << endl;
    return 0;
}

and I return this:

Input a sentence:

hello

h: 1

e: 1

l: 2

l: 2

o: 1

so I kind of have something but I can't figure out how to make the letters not repeat more than once. I know that I need to store them in an array but it's out of my ken.

Answer 1

You were very close, nice try! I liked the approach with the counters array, where every cell would represent the frequency of a letter in the given string.

So, just go and update this array, as this answer implies How to get character's position in alphabet in C language? , without the plus one they mention there, since you want the index of the letter in your array. In other words, for 'a', you need 0, 'b', you need 1 and so on.

Then, in the printing phase, just use the above link's suggestion in the reverse way. When you print the i-th non-zero element of counters , print the i-th element of the element, which will reveal the letter in question.

Putting all together, you get:

#include <iostream>
#include <ctime>
#include <cstdlib>
#include <cmath>
#include <string>

using namespace std;

int main()
{
    string st = "";

    cout << "Input a sentence: " << endl;
    getline(cin, st);
    int index = 0;
    int index2 = 0;
    int counters[26] = {0};
    for(size_t i = 0; i < st.length(); i++)
    {
        int counter = 0;
        index = st.find(st[i],0);
        for(size_t j = 0; j < st.length(); j++)
        {
            index2 = st.find(st[j]);
            if(index == index2)
            {
                counter++;
            }
        }
        counters[st[i] - 'a'] = counter; // update 'counters' array
    }
    for(int i = 0; i < 26; ++i)
        if(counters[i] != 0)            // print non-zero counters 
            cout << (char)(i + 'a') << ": " << counters[i] << endl;
    return 0;
}

Output:

e: 1
h: 1
l: 2
o: 1

Answer 2

I would do something like this:

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

int main()
{
    std::string st;

    std::cout << "Input a sentence: " << std::endl;
    std::getline(std::cin, st);

    std::map<char, int> m;

    for (const char c : st)
        ++m[c];

    for (const std::pair<char, int>& me : m)
        std::cout << me.first << ": " << me.second << std::endl;
}

Answer 3

lechuga2000 beat me to the post, but this is my suggestion:

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

int main()
{
    std::string input_sentence = "Now is the time for all good men to come to the aid of the party.";

/*
    std::cout << "Input a sentence: " << std::endl;
    std::getline(std::cin, input_sentence);
*/

    std::map<char, int> character_counts;

    for (const auto character : input_sentence)
        ++character_counts[character];

    for (const auto counted_character : character_counts)
        std::cout << counted_character.first << ": " << counted_character.second << '\n';

    return 0;
}

And here is the output:

 : 15
.: 1
N: 1
a: 3
c: 1
d: 2
e: 6
f: 2
g: 1
h: 3
i: 3
l: 2
m: 3
n: 1
o: 8
p: 1
r: 2
s: 1
t: 7
w: 1
y: 1