给定N个整数的数组arr [] 。任务是根据元素的频率以降序对数组arr []进行排序。
注意：如果两个元素的频率相同，则较小的元素应首先出现。
例子：
Input: arr[] = { 4, 4, 5, 6, 4, 2, 2, 8, 5 }
Output: 4 4 4 2 2 5 5 6 8
Input: arr[] = { 9, 9, 5, 8, 5 }
Output: 5 5 9 9 8
方法：
- 将所有元素的频率存储在数组arr []中。
- 对于arr [] = {4，4，5，6，4，2，2，8，5}
上述阵列的频率阵列为：
freq [] = {0，0，2，0，3，2，0，1，0，1} - 遍历频率数组，并对所有频率大于1的元素，将数组arr []中的值更新为：
freq[2] = 2
arr[0] = 100000*freq[2] + (100000 – 2) = 299998
freq[4] = 3
arr[1] = 100000*freq[2] + (100000 – 4) = 399996
freq[5] = 2
arr[2] = 100000*freq[2] + (100000 – 5) = 299995
freq[6] = 2
arr[3] = 100000*freq[2] + (100000 – 6) = 199994
freq[8] = 2
arr[4] = 100000*freq[2] + (100000 – 2) = 199994
Now array becomes:
arr[] = {299998, 399996, 299995, 199994, 199992, 2, 2, 8, 5}
- 以降序对数组arr []进行排序。
- 遍历数组arr []并按照步骤2中数组元素的更新获取该元素的元素和频率，请执行以下操作：
To get the frequency of current element:
frequency = arr[i]/100000;
To get the value:
value = 100000 – arr[i]%100000
例如：
if arr[i] = 399996
frequency = arr[i]/100000 = 399996/100000 = 3
value = 100000 – arr[i]%100000 = 100000 – 99996 = 4
The element 4 is having frequency 3.
- 对于arr []中的每个元素，找到每个索引处的值和频率(例如f )，然后将值f打印多次。
下面是上述方法的实现：
C++
// C++ program to sort an array in
// decreasing order of their frequnecy
#include "bits/stdc++.h"
using namespace std;
// Function that return the index
// upto all the array elements are
// updated.
int sortByFreq(int* arr, int n)
{
// Initialise maxE = -1
int maxE = -1;
// Find the maximum element of
// arr[]
for (int i = 0; i < n; i++) {
maxE = max(maxE, arr[i]);
}
// Create frequency array freq[]
int freq[maxE + 1] = { 0 };
// Update the frequency array as
// per the occurence of element in
// arr[]
for (int i = 0; i < n; i++) {
freq[arr[i]]++;
}
// Initialise cnt to 0
int cnt = 0;
// Traversing freq[]
for (int i = 0; i <= maxE; i++) {
// If freq of an element is
// greater than 0 update the
// value of arr[] at index cnt
// & increment cnt
if (freq[i] > 0) {
int value = 100000 - i;
arr[cnt] = 100000 * freq[i] + value;
cnt++;
}
}
// Return cnt
return cnt;
}
// Function that print array arr[]
// elements in sorted order
void printSortedArray(int* arr, int cnt)
{
// Traversing arr[] till index cnt
for (int i = 0; i < cnt; i++) {
// Find frequency of elements
int frequency = arr[i] / 100000;
// Find value at index i
int value = 100000 - (arr[i] % 100000);
// Traversing till frequency
// to print value at index i
for (int j = 0; j < frequency; j++) {
cout << value << ' ';
}
}
}
// Driver code
int main()
{
int arr[] = { 4, 4, 5, 6, 4, 2, 2, 8, 5 };
// Size of array arr[]
int n = sizeof(arr) / sizeof(arr[0]);
// Function call to get cnt
int cnt = sortByFreq(arr, n);
// Sort the arr[] in decreasing order
sort(arr, arr + cnt, greater());
// Function that prints elements
// in decreasing order
printSortedArray(arr, cnt);
return 0;
} Java
// Java program to sort an array in
// decreasing order of their frequnecy
import java.util.*;
class GFG{
// Function that return the index
// upto all the array elements are
// updated.
static int sortByFreq(Integer []arr, int n)
{
// Initialise maxE = -1
int maxE = -1;
// Find the maximum element of
// arr[]
for (int i = 0; i < n; i++) {
maxE = Math.max(maxE, arr[i]);
}
// Create frequency array freq[]
int freq[] = new int[maxE + 1];
// Update the frequency array as
// per the occurence of element in
// arr[]
for (int i = 0; i < n; i++) {
freq[arr[i]]++;
}
// Initialise cnt to 0
int cnt = 0;
// Traversing freq[]
for (int i = 0; i <= maxE; i++) {
// If freq of an element is
// greater than 0 update the
// value of arr[] at index cnt
// & increment cnt
if (freq[i] > 0) {
int value = 100000 - i;
arr[cnt] = 100000 * freq[i] + value;
cnt++;
}
}
// Return cnt
return cnt;
}
// Function that print array arr[]
// elements in sorted order
static void printSortedArray(Integer []arr, int cnt)
{
// Traversing arr[] till index cnt
for (int i = 0; i < cnt; i++) {
// Find frequency of elements
int frequency = arr[i] / 100000;
// Find value at index i
int value = 100000 - (arr[i] % 100000);
// Traversing till frequency
// to print value at index i
for (int j = 0; j < frequency; j++) {
System.out.print(value + " ");
}
}
}
// Driver code
public static void main(String[] args)
{
Integer arr[] = { 4, 4, 5, 6, 4, 2, 2, 8, 5 };
// Size of array arr[]
int n = arr.length;
// Function call to get cnt
int cnt = sortByFreq(arr, n);
// Sort the arr[] in decreasing order
Arrays.sort(arr, Collections.reverseOrder());
// Function that prints elements
// in decreasing order
printSortedArray(arr, cnt);
}
}
// This code is contributed by sapnasingh4991Python3
# Python program to sort an array in
# decreasing order of their frequnecy
# Function that return the index
# upto all the array elements are
# updated.
def sortByFreq(arr, n):
# Initialise maxE = -1
maxE = -1;
# Find the maximum element of
# arr[]
for i in range(n):
maxE = max(maxE, arr[i])
# Create frequency array freq[]
freq = [0]*(maxE + 1);
# Update the frequency array as
# per the occurence of element in
# arr[]
for i in range(n):
freq[arr[i]] += 1;
# Initialise cnt to 0
cnt = 0;
# Traversing freq[]
for i in range(maxE+1):
# If freq of an element is
# greater than 0 update the
# value of arr[] at index cnt
# & increment cnt
if (freq[i] > 0):
value = 100000 - i;
arr[cnt] = 100000 * freq[i] + value;
cnt += 1;
# Return cnt
return cnt;
# Function that print array arr[]
# elements in sorted order
def printSortedArray(arr, cnt):
# Traversing arr[] till index cnt
for i in range(cnt):
# Find frequency of elements
frequency = arr[i] / 100000;
# Find value at index i
value = 100000 - (arr[i] % 100000);
# Traversing till frequency
# to print value at index i
for j in range(int(frequency)):
print(value, end=" ")
# Driver code
if __name__=='__main__':
arr = [ 4, 4, 5, 6, 4, 2, 2, 8, 5 ]
# Size of array arr[]
n = len(arr)
# Function call to get cnt
cnt = sortByFreq(arr, n);
# Sort the arr[] in decreasing order
arr.sort(reverse = True)
# Function that prints elements
# in decreasing order
printSortedArray(arr, cnt);
# This code is contributed by Princi SinghC#
// C# program to sort an array in
// decreasing order of their frequnecy
using System;
class GFG {
// Function that return the index
// upto all the array elements are
// updated.
static int sortByFreq(int[] arr, int n)
{
// Initialise maxE = -1
int maxE = -1;
// Find the maximum element of
// arr[]
for (int i = 0; i < n; i++)
{
maxE = Math.Max(maxE, arr[i]);
}
// Create frequency array freq[]
int[] freq = new int[maxE + 1];
// Update the frequency array as
// per the occurence of element in
// arr[]
for (int i = 0; i < n; i++)
{
freq[arr[i]]++;
}
// Initialise cnt to 0
int cnt = 0;
// Traversing freq[]
for (int i = 0; i <= maxE; i++)
{
// If freq of an element is
// greater than 0 update the
// value of arr[] at index cnt
// & increment cnt
if (freq[i] > 0)
{
int value = 100000 - i;
arr[cnt] = 100000 * freq[i] + value;
cnt++;
}
}
// Return cnt
return cnt;
}
// Function that print array arr[]
// elements in sorted order
static void printSortedArray(int[] arr, int cnt)
{
// Traversing arr[] till index cnt
for (int i = 0; i < cnt; i++)
{
// Find frequency of elements
int frequency = arr[i] / 100000;
// Find value at index i
int value = 100000 - (arr[i] % 100000);
// Traversing till frequency
// to print value at index i
for (int j = 0; j < frequency; j++)
{
Console.Write(value + " ");
}
}
}
// Driver code
public static void Main()
{
int[] arr = { 4, 4, 5, 6, 4, 2, 2, 8, 5 };
// Size of array arr[]
int n = arr.Length;
// Function call to get cnt
int cnt = sortByFreq(arr, n);
// Sort the arr[] in decreasing order
Array.Sort(arr);
Array.Reverse(arr);
// Function that prints elements
// in decreasing order
printSortedArray(arr, cnt);
}
}
// This code is contributed by subhamahato348输出：
4 4 4 2 2 5 5 6 8时间复杂度： O(N * log N)
辅助空间： O(N)