因此下面的代码具有O(N ^ 2)的复杂度,是否有一种简单的方法来编辑此代码以提供O(n ^(3/2))和O(n ^(4 / 3))。这里有一些提示:https://en.wikipedia.org/wiki/Shellsort#Gap_sequences,但是在实现它方面遇到了问题。 (我的任务是比较图中这三个算法的速度)
#include <iostream>
using namespace std;
int shellSort(int arr[], int n)
{
for (int gap = n/2; gap > 0; gap /= 2)
{
for (int i = gap; i < n; i += 1)
{
int temp = arr[i];
int j;
for (j = i; j >= gap && arr[j - gap] > temp; j -= gap)
arr[j] = arr[j - gap];
arr[j] = temp;
}
}
return 0;
}
void printArray(int arr[], int n)
{
for (int i=0; i<n; i++)
cout << arr[i] << " ";
}
int main()
{
int arr[] = {12, 34, 54, 2, 3}, i;
int n = sizeof(arr)/sizeof(arr[0]);
cout << "Array before sorting: \n";
printArray(arr, n);
shellSort(arr, n);
cout << "\nArray after sorting: \n";
printArray(arr, n);
return 0;
}