C ++でO（n）時間未満で、制限された範囲の配列内の各要素の頻度を見つけます

例

#include<iostream>
#include<vector>
using namespace std;
void calculateFreq(int arr[], int left, int right, vector<int>& frequency) {
   if (arr[left] == arr[right])
      frequency[arr[left]] += right - left + 1;
   else {
      int mid = (left + right) / 2;
      calculateFreq(arr, left, mid, frequency);
      calculateFreq(arr, mid + 1, right, frequency);
   }
}
void getAllFrequency(int arr[], int n) {
   vector<int> frequency(arr[n - 1] + 1, 0);
   calculateFreq(arr, 0, n - 1, frequency);
   for (int i = 0; i <= arr[n - 1]; i++)
      if (frequency[i] != 0)
         cout << "Frequency of element " << i << " is " << frequency[i] << endl;
}
int main() {
   int arr[] = { 10, 10, 10, 20, 30, 30, 50, 50, 80, 80, 80, 90, 90, 99 };
   int n = sizeof(arr) / sizeof(arr[0]);
   getAllFrequency(arr, n);
}

出力

Frequency of element 10 is 3
Frequency of element 20 is 1
Frequency of element 30 is 2
Frequency of element 50 is 2
Frequency of element 80 is 3
Frequency of element 90 is 2
Frequency of element 99 is 1