根据元素在 0 到 n-1 范围内的频率填充数组

给定一个允许重复的正整数数组。该数组包含范围从 0 到 n-1 的元素。任务是填充数组，使 arr[i] 包含 i 的频率。

例子 ：

Input  : arr[] = {1, 4, 3, 4, 1, 1, 4, 4, 4, 7}
Output : arr[] = {0, 3, 0, 1, 5, 0, 0, 1, 0, 0}
Here 0 appears 0 times, so arr[0] is 0
     1 appears 3 times 
     2 appears 0 times 
     3 appears 0 times 
     4 appears 5 times 
     ..........

Input  : arr[] = {1, 2, 3, 4, 1, 1, 4, 5, 6, 7}
Output : arr[] = {0, 3, 1, 1, 2, 1, 1, 1, 0, 0} 

这个问题的一个简单解决方案是运行两个循环。外循环一一挑选元素。内部循环计算拾取元素的频率并将频率存储在最终数组中。

一个有效的解决方案是使用大小为 n 的辅助数组。

1) Create an array temp[0..n-1] and 
   initialize it as 0.
2) Traverse given array and do following
   for every element arr[i].
     temp[arr[i]]++
3) Copy temp[] to arr[].

下面是上述步骤的实现。

C++

// C++ program to fill an array with frequencies.
#include
using namespace std;
 
// Fills arr[] with frequencies of elements
void fillWithFreq(int arr[], int n)
{
    int temp[n];
    memset(temp, 0, sizeof(temp));
 
    // Fill temp with frequencies
    for (int i=0; i

---

Java

// Java program to fill an array with frequencies.
import java.util.Arrays;
 
class GFG {
     
    // Fills arr[] with frequencies of elements
    static void fillWithFreq(int arr[], int n)
    {
         
        int temp[]=new int[n];
        Arrays.fill(temp, 0);
     
        // Fill temp with frequencies
        for (int i = 0; i < n; i++)
            temp[arr[i]] += 1;
     
        // Copy temp to array
        for (int i = 0; i < n; i++)
            arr[i] = temp[i];
    }
     
    // Driver method
    public static void main(String[] args)
    {
         
        int arr[] = {5, 2, 3, 4, 5, 5, 4, 5, 6, 7};
        int n = arr.length;
         
        fillWithFreq(arr, n);
         
        for (int i=0; i

---

Python3

# Python3 program to fill an
# array with frequencies.
 
# Fills arr[] with frequencies of elements
def fillWithFreq(arr, n):
 
    temp = [0 for i in range(n)]
 
    # Fill temp with frequencies
    for i in range(n):
        temp[arr[i]] += 1
 
    # Copy temp to array
    for i in range(n):
        arr[i] = temp[i]
 
# Driver Code
arr = [5, 2, 3, 4, 5, 5, 4, 5, 6, 7]
n = len(arr)
fillWithFreq(arr, n)
for i in range(n):
    print(arr[i], end = " ")
     
# This code is contributed by Anant Agarwal.

C#

// C# program to fill an
// array with frequencies.
using System;
 
class GFG {
     
    // Fills arr[] with frequencies of elements
    static void fillWithFreq(int []arr, int n)
    {
        int []temp = new int[n];
        for(int i = 0; i < n; i++)
         temp[i] = 0;
          
        // Fill temp with frequencies
        for (int i = 0; i < n; i++)
            temp[arr[i]] += 1;
     
        // Copy temp to array
        for (int i = 0; i < n; i++)
            arr[i] = temp[i];
    }
     
    // Driver method
    public static void Main()
    {
        int []arr = {5, 2, 3, 4, 5,
                     5, 4, 5, 6, 7};
        int n = arr.Length;
         
        // Function calling
        fillWithFreq(arr, n);
         
        for (int i = 0; i < n; i++)
        Console.Write(arr[i] + " ");
    }
}
 
// This code is contributed by nitin mittal.

Javascript

    // Javascript program to fill an
    // array with frequencies.
     
    // Fills arr[] with frequencies of elements
    function fillWithFreq(arr, n)
    {
        let temp = new Array(n);
        for(let i = 0; i < n; i++)
         temp[i] = 0;
            
        // Fill temp with frequencies
        for (let i = 0; i < n; i++)
            temp[arr[i]] += 1;
       
        // Copy temp to array
        for (let i = 0; i < n; i++)
            arr[i] = temp[i];
    }
     
    let arr = [5, 2, 3, 4, 5, 5, 4, 5, 6, 7];
    let n = arr.length;
 
    // Function calling
    fillWithFreq(arr, n);
 
    for (let i = 0; i < n; i++)
      document.write(arr[i] + " ");
     

输出 ：

0 0 1 1 2 4 1 1 0 0 

时间复杂度： O(n)
辅助空间： O(n)