计算C ++中相同数组值的出现次数

Question

I recently have been building a program where:

1. A user is asked to enter a number that will represent the size of a character array.
2. Then they are asked whether they will want the program to fill the values automatically, or they could press M so they could enter the values manually. They may only enter a-zA-Z values, or they will see an error.
3. At the end of the program, I am required to count every duplicate value and display it, for example:

An array of 5 characters consists of A;A;A;F;G;

The output should be something like:

A - 3

F - 1

G - 1

I could do this easily, however, the teacher said I may not use an additional array, but I could make a good use of a few more variables and I also can't use a switch element. I'm totally lost and I can't find a solution. I've added the code down below. I have done everything, but the counting part.

#pragma hdrstop
#pragma argsused

#include <tchar.h>
#include <iostream.h>
#include <conio.h>
#include <math.h>
#include <stdio.h>
#include <time.h>
#include <ctype.h>

void main() {
int n, i = 0;
char masiva_izvele, array[100], masiva_burts;
cout << "Enter array size: ";
cin >> n;

clrscr();

cout << "You chose an array of size " << n << endl;
cout << "\nPress M to enter array values manually\nPress A so the program could do it for you.\n\n";

cout << "Your choice (M/A): ";
cin >> masiva_izvele;

if (masiva_izvele == 'M' || masiva_izvele == 'm') {
    clrscr();
    for (i = 0; i < n; i++) {
        do {
            cout << "Enter " << i + 1 << " array element: ";
            flushall();
            cin >> masiva_burts;
            cout << endl << int(masiva_burts);
            if (isalpha(masiva_burts)) {
                clrscr();
                array[i] = masiva_burts;

            }
            else {
                clrscr();
                cout << "Unacceptable value, please enter a value from the alphabet!\n\n";
            }
        }
        while (!isalpha(masiva_burts));
    }


}
else if (masiva_izvele == 'A' || masiva_izvele == 'a') {
    clrscr();
    for (i = 0; i < n; i++) {

        array[i] = rand() % 25 + 65;

    }
}

clrscr();

cout << "Masivs ir izveidots! \nArray size is " << n <<
    "\nArray consists of following elements:\n\n";

for (i = 0; i < n; i++) {

    cout << array[i] << "\t";

}         



cout << "\n\nPress any key to view the amount of every element in array.";

//The whole thing I miss goes here, teacher said I would need two for loops but I can't seem to find the solution.

getch();

}

I would be very thankful for a solution so I could move on and forgive my C++ amateur-ness as I've picked this language up just a few days ago.

Thanks.

EDIT: Edited title to suit the actual problem, as suggested in comments.

Answer 1

One possible way is to sort the array, and then iterate over it counting the current letter. When the letter changes (for example from 'A' to 'F' as in your example) print the letter and the count. Reset the counter and continue counting the next character.

Answer 2

The main loop should run foreach character in your string.

The secondary loop should run each time the main "passes by" to check if the current letter is in array. If it's there, then ++.

Answer 3

Assuming upper and lower case letters are considered to be equal (otherwise, you need an array twice the size as the one proposed:

std::array<unsigned int, 26> counts; //!!!

// get number of characters to read

for(unsigned int i = 0; i < charactersToRead; ++i)
{
    char c; // get a random value or read from console
    // range check, calculate value in range [0; 25] from letter...

    // now the trick: just do not store the value in an array,
    // evaluate it  d i r e c t l y  instead:
    ++counts[c]; 
}

// no  a d d i t i o n a l  array so far...

char c = 'a';
for(auto n : counts)
{
    if(n > 0) // this can happen now...
    {
        // output c and n appropriately! 
    }
    ++c; // only works on character sets without gaps in between [a; z]!
         // special handling required if upper and lower case not considered equal!
}

Side note: (see CiaPan's comment to the question): If only true duplicates to be counted, must be if(n > 1) within last loop!

Answer 4

Add the array char chars[52] and count chars in this array. Then print out chars corresponding to the array, which count is more than 1.

std::unordered_map<char, int> chars;
...
char c = ...;
if ('A' <= c && c <= 'Z')
  ++chars[c];
else if ('a' <= c && c <= 'z')
  ++chars[c];
else
  // unexpected char
...
for (const auto p : chars)
  std::cout << p.first << ": " << p.second << " ";