在C ++中以小于O(n)的时间在有限范围的数组中查找每个元素的频率
假设我们有一个整数数组。数组为A，大小为n。我们的任务是找到阵列中所有元素的频率小于O(n)时间。元素的大小必须小于一个等于M的值。在这里，我们将使用二进制搜索方法。在这里，如果末端元素不同，则将数组递归分为两部分；如果两个末端元素相同，则意味着数组中的所有元素与已排序的数组相同。
示例
#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);
}输出结果
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