Zählen mehrere Array-Elemente

Question

ich ein C++ Programm geschrieben, das mir die Anzahl der Wiederholungen der Array-Elemente zeigt ... mein Quellcode ist:

#include <iostream> 
#include <string> 
using namespace std; 
int main() { 
    int x[20] = {1,1,32,43,54,65,76,76,76,2,12,12,32,43,54,3,3,23,1,43}; 
    for (int i=0;i<20;i++) { 
    } 

    for (int i=0;i<20;i++) { 
     int count=1; 
     for (int j=i+1;j<19;j++) { 
      if (x[i]==x[j]) count++; 
     } 
     cout<<"The number "<<x[i]<<" is repeated "<<count<<" times"<<"\n"; 
    } 
} 

und aus diesem heraus genommen wird:

The number 1 is repeated 3 times 
The number 1 is repeated 2 times 
The number 32 is repeated 2 times 
The number 43 is repeated 2 times 
The number 54 is repeated 2 times 
The number 65 is repeated 1 times 
The number 76 is repeated 3 times 
The number 76 is repeated 2 times 
The number 76 is repeated 1 times 
The number 2 is repeated 1 times 
The number 12 is repeated 2 times 
The number 12 is repeated 1 times 
The number 32 is repeated 1 times 
The number 43 is repeated 1 times 
The number 54 is repeated 1 times 
The number 3 is repeated 2 times 
The number 3 is repeated 1 times 
The number 23 is repeated 1 times 
The number 1 is repeated 1 times 
The number 43 is repeated 1 times 

Das Problem ist Ausgabe zeigt Array-Element jedes Mal, aber ich möchte, dass mein Programm zeigt nur die wiederholte Array nur einmal. und ich will kein neues Array definieren. Jeder hat eine Ahnung was los ist ??

Hinweis: dass ohne Definition jede neue Anordnung und ohne Programmausgabesortierung sollte wie folgt sein:

The number 1 is repeated 3 times 
    The number 32 is repeated 2 times 
    The number 43 is repeated 3 times 
    The number 54 is repeated 2 times 
    The number 65 is repeated 1 times 
    The number 76 is repeated 3 times 
    The number 2 is repeated 1 times 
    The number 12 is repeated 2 times 
    The number 3 is repeated 2 times 
    The number 23 is repeated 1 times 

Answer 1

Sie eine Karte verwenden kann, um Elemente zu zählen, die Ihre Anforderung erfüllen, dass Sie schaffen keine neue Array.

std::map<int, int> counts; 
for(auto&& elem : x) 
{ 
    counts[elem]++; 
} 

for(auto&& item : counts) 
{ 
    std::cout << "The number " << item.first << " is repeated " << item.second << " times; 
} 

Answer 2

Ohne zusätzliche Strukturen, können Sie einfach tun:

#include <algorithm> 
#include <iostream> 
#include <string> 

int main() { 
    const int x[20] = {1,1,32,43,54,65,76,76,76,2,12,12,32,43,54,3,3,23,1,43}; 

    for (int i=0;i<20;i++) { 
     if (std::find(x, x + i, x[i]) != x + i) { 
      continue; 
     } 
     const auto count = std::count(x + i, x + 20, x[i]); 
     std::cout << "The number " << x[i] << " is repeated " << count << " times\n"; 
    } 
} 

Demo

Answer 3

Dies ist ein Betrieb zu reduzieren. C++ hat in Algorithmen eine linksseitige Operation namens std::accumulate. Füttern Sie es mit einem std::map, um einen Count-Record zu erstellen.

auto count_map = std::accumulate(std::begin(x), std::end(x), 
           std::map<int, int>{}, 
           [] (auto& m, int val) -> decltype(m) { 
    ++m[val]; 
    return m; 
}); 

// Output result 
for (auto&& [val, count] : count_map) { 
    std::cout << "Value: " << val << " - Count: " << count << std::endl; 
} 

# Output: 
Value: 1 - Count: 3 
Value: 2 - Count: 1 
Value: 3 - Count: 2 
Value: 12 - Count: 2 
Value: 23 - Count: 1 
Value: 32 - Count: 2 
Value: 43 - Count: 3 
Value: 54 - Count: 2 
Value: 65 - Count: 1 
Value: 76 - Count: 3