以 O(n) 复杂度实现少于 100 个数字的排序的 C++ 程序

2023-06-20 12:15:03 390

计数排序技术的复杂性

时间复杂度：O(n+r)

空间复杂度：O(n+r)

Input − A list of unsorted data: 2 5 6 2 3 10 3 6 7 8Output − Array after Sorting: 2 2 3 3 5 6 6 7 8 10

算法

countingSort(array,size)

输入-数据数组，以及数组中的总数

输出-排序后的数组

Begin
   max = get maximum element from array.
   define count array of size [max+1]
   for i := 0 to max do
      count[i] = 0 //将计数数组中的所有元素设置为0
   done
   for i := 1 to size do
      increase count of each number which have found in the array
   done
   for i := 1 to max do
      count[i] = count[i] + count[i + 1] //找到累积频率
   done
   for i := size to 1 decrease by 1 do
      store the number in the output array
      decrease count[i]
   done
   return the output array
End

示例代码

#include 
using namespace std;
void counting_sort(int array[], int n) {
   int i, j;
   int count[n];
   for (i = 0; i < n; i++)
      count[i] = 0;
   for (i = 0; i < n; i++)
      (count[array[i]])++;
   for (i = 0, j = 0; i < n; i++)
      for (; count[i] > 0; (count[i])--)
         array[j++] = i;
}
int main() {
   int array[100], i, num;
   cout << "输入数组的大小： ";
   cin >> num;
   cout << "输入 " << num << " elements to be sorted:" << endl;
   for (i = 0; i < num; i++)
      cin >> array[i];
   cout << "\nThe array of elements before sorting : " <输出结果输入数组的大小： 8

输入 8 elements to be sorted:
54 89 23 20 18 88 65 31

The array of elements before sorting :
54 89 23 20 18 88 65 31

The array of elements after sorting :
54 89 23 20 18 88 65 31

以 O(n) 复杂度实现少于 100 个数字的排序的 C++ 程序

计数排序技术的复杂性

算法

countingSort(array,size)

示例代码

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