按降序打印数字及其频率

给定一个数组arr ，任务是按照降序打印数组的元素及其频率。
例子：

Input: arr[] = {1, 3, 3, 3, 4, 4, 5} 
Output: 5 occurs 1 times 
4 occurs 2 times 
3 occurs 3 times 
1 occurs 1 times
Input: arr[] = {1, 1, 1, 2, 3, 4, 9, 9, 10} 
Output: 10 occurs 1 times 
9 occurs 2 times 
4 occurs 1 times 
3 occurs 1 times 
2 occurs 1 times 
1 occurs 3 times 
 

幼稚的方法：使用某些数据结构(例如多集)以递减的顺序存储元素，然后按计数顺序逐个打印元素，然后从数据结构中删除它们。对于使用的数据结构，时间复杂度将为O(N log N)，辅助空间将为O(N)。
下面是上述方法的实现：

CPP

// C++ program to print the elements in
// descending along with their frequencies
#include 
using namespace std;
 
// Function to print the elements in descending
// along with their frequencies
void printElements(int a[], int n)
{
 
    // A multiset to store elements in decreasing order
    multiset > ms;
 
    // Insert elements in the multiset
    for (int i = 0; i < n; i++) {
        ms.insert(a[i]);
    }
 
    // Print the elements along with their frequencies
    while (!ms.empty()) {
 
        // Find the maximum element
        int maxel = *ms.begin();
 
        // Number of times it occurs
        int times = ms.count(maxel);
 
        cout << maxel << " occurs " << times << " times\n";
 
        // Erase the maxel
        ms.erase(maxel);
    }
}
 
// Driver Code
int main()
{
    int a[] = { 1, 1, 1, 2, 3, 4, 9, 9, 10 };
    int n = sizeof(a) / sizeof(a[0]);
    printElements(a, n);
    return 0;
}

输出：

10 occurs 1 times
9 occurs 2 times
4 occurs 1 times
3 occurs 1 times
2 occurs 1 times
1 occurs 3 times

高效方法：按降序对数组进行排序，然后从头开始打印元素及其频率。
下面是上述方法的实现：

C++

// C++ program to print the elements in
// descending along with their frequencies
#include 
using namespace std;
 
// Function to print the elements in descending
// along with their frequencies
void printElements(int a[], int n)
{
 
    // Sorts the element in decreasing order
    sort(a, a + n, greater());
    int cnt = 1;
 
    // traverse the array elements
    for (int i = 0; i < n - 1; i++) {
 
        // Prints the number and count
        if (a[i] != a[i + 1]) {
            cout << a[i] << " occurs " << cnt << " times\n";
            cnt = 1;
        }
        else
            cnt += 1;
    }
 
    // Prints the last step
    cout << a[n - 1] << " occurs " << cnt << " times\n";
}
 
// Driver Code
int main()
{
    int a[] = { 1, 1, 1, 2, 3, 4, 9, 9, 10 };
    int n = sizeof(a) / sizeof(a[0]);
 
    printElements(a, n);
    return 0;
}

Java

// Java program to print the elements in
// descending along with their frequencies
import java.util.*;
 
class GFG
{
 
// Function to print the elements in descending
// along with their frequencies
static void printElements(int a[], int n)
{
 
    // Sorts the element in decreasing order
    Arrays.sort(a);
    a = reverse(a);
    int cnt = 1;
 
    // traverse the array elements
    for (int i = 0; i < n - 1; i++)
    {
 
        // Prints the number and count
        if (a[i] != a[i + 1])
        {
            System.out.print(a[i]+ " occurs " +
                            cnt + " times\n");
            cnt = 1;
        }
        else
            cnt += 1;
    }
 
    // Prints the last step
    System.out.print(a[n - 1]+ " occurs " +
                    cnt + " times\n");
}
 
static int[] reverse(int a[])
{
    int i, n = a.length, t;
    for (i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
}
 
// Driver Code
public static void main(String[] args)
{
    int a[] = { 1, 1, 1, 2, 3, 4, 9, 9, 10 };
    int n = a.length;
 
    printElements(a, n);
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to print the elements in
# descending along with their frequencies
 
# Function to print the elements in
# descending along with their frequencies
def printElements(a, n) :
 
    # Sorts the element in decreasing order
    a.sort(reverse = True)
    cnt = 1
 
    # traverse the array elements
    for i in range(n - 1) :
 
        # Prints the number and count
        if (a[i] != a[i + 1]) :
            print(a[i], " occurs ", cnt, "times")
            cnt = 1
         
        else :
            cnt += 1
     
    # Prints the last step
    print(a[n - 1], "occurs", cnt, "times")
 
# Driver Code
if __name__ == "__main__" :
 
    a = [ 1, 1, 1, 2,
          3, 4, 9, 9, 10 ]
    n = len(a)
 
    printElements(a, n)
     
# This code is contributed by Ryuga

C＃

// C# program to print the elements in
// descending along with their frequencies
using System;
 
class GFG
{
 
// Function to print the elements in descending
// along with their frequencies
static void printElements(int []a, int n)
{
 
    // Sorts the element in decreasing order
    Array.Sort(a);
    a = reverse(a);
    int cnt = 1;
 
    // traverse the array elements
    for (int i = 0; i < n - 1; i++)
    {
 
        // Prints the number and count
        if (a[i] != a[i + 1])
        {
            Console.Write(a[i]+ " occurs " +
                            cnt + " times\n");
            cnt = 1;
        }
        else
            cnt += 1;
    }
 
    // Prints the last step
    Console.Write(a[n - 1]+ " occurs " +
                    cnt + " times\n");
}
 
static int[] reverse(int []a)
{
    int i, n = a.Length, t;
    for (i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []a = { 1, 1, 1, 2, 3, 4, 9, 9, 10 };
    int n = a.Length;
 
    printElements(a, n);
}
}
 
// This code is contributed by PrinciRaj1992

Java脚本

 
// javascript program to print the elements in
// descending along with their frequencies
 
   
// Function to print the elements in descending
// along with their frequencies
  function printElements(a,  n)
   
{
   
    // Sorts the element in decreasing order
    a=a.sort(compare);
    a = reverse(a);
    var cnt = 1;
   
    // traverse the array elements
    for (var i = 0; i < n - 1; i++)
    {
   
        // Prints the number and count
        if (a[i] != a[i + 1])
        {
            document.write(a[i]+ " occurs " +   cnt + " times" + "<br>");
            cnt = 1;
        }
        else
            cnt += 1;
    }
   
    // Prints the last step
    document.write(a[n - 1]+ " occurs " + cnt + " times" + "<br>");
}
   
  function reverse(a){
    var i, n = a.length, t;
     
    for (i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
}
 
function compare(a, b) {
    if (a < b) {
        return -1;
    } else if (a > b) {
        return 1;
    } else {
        return 0;
    }
}
   
// Driver Code
 
    var a = [ 1, 1, 1, 2, 3, 4, 9, 9, 10 ];
    var n = a.length;
   
    printElements(a, n);
  
 // This code is contributed by bunnyram19.

输出：

10 occurs 1 times
9 occurs 2 times
4 occurs 1 times
3 occurs 1 times
2 occurs 1 times
1 occurs 3 times

想要从精选的最佳视频中学习并解决问题，请查看有关从基础到高级C++的C++基础课程以及有关语言和STL的C++ STL课程。要完成从学习语言到DS Algo等的更多准备工作，请参阅“完整面试准备课程” 。