Question

我使用mergesort编写反转计数代码，但因为输入是一个大的const

int数组，如何让它更快？（让o（nlogn）的系数变小？）

是否有一些代码提示可以做到？

提前谢谢。

#include<iostream>
using namespace std;
int invCount(int*, int);
int merge(int*, int*, int, int*, int);
int NumberOfInversion(const int*, int);

int main(void){
    const int array[] = {0, 1, 4, 3, 2};
    cout << NumberOfInversion(array, 5) << "times" << endl;
    return 0;
}

    int invCount(int *array, int length){

        if(length <= 1)return 0;

        int middle = (length + 1) / 2;

        int left[middle];
        int right[length - middle];


        for(int i = 0; i < middle; i ++)left[i] = array[i];
        for(int i = middle; i < length; i ++)right[i - middle] = array[i];

        return invCount(left, middle) + invCount(right, length - middle
        ) + merge(array, left, middle, right, length - middle);
    }


    int merge(int* array, int* left, int leftLength, int* right, int rightLength){
       int i = 0, j = 0, count = 0;

       while(i < leftLength && j < rightLength){
           if (left[i] <= right[j]){
               array[i + j] = left[i];
               i ++;
           }
           else {
               array[i + j] = right[j];
               j ++;
               count += leftLength - i;
           }
       }
       if(i == leftLength){
            while(j < rightLength){
                array[i + j] = right[j];
                j ++;
            }
       }
       else{
            while(i < leftLength){
                array[i + j] = left[i];
                i ++;
            }
        }
       return count;
    }


    int NumberOfInversion(const int *A, int N)
    {
        int input[N];
        for(int i = 0; i < N; i ++)input[i] = A[i];
        int result = invCount(input, N);
        return result;
    }

ps：所有可能的对中约有20％被反转

Answer 1

以下是计数反转的C代码

#include <stdio.h>
#include <stdlib.h>

int  _mergeSort(int arr[], int temp[], int left, int right);
int merge(int arr[], int temp[], int left, int mid, int right);

/* This function sorts the input array and returns the
   number of inversions in the array */
int mergeSort(int arr[], int array_size)
{
    int *temp = (int *)malloc(sizeof(int)*array_size);
    return _mergeSort(arr, temp, 0, array_size - 1);
}

/* An auxiliary recursive function that sorts the input array and
  returns the number of inversions in the array. */
int _mergeSort(int arr[], int temp[], int left, int right)
{
  int mid, inv_count = 0;
  if (right > left)
  {
    /* Divide the array into two parts and call _mergeSortAndCountInv()
       for each of the parts */
    mid = (right + left)/2;

    /* Inversion count will be sum of inversions in left-part, right-part
      and number of inversions in merging */
    inv_count  = _mergeSort(arr, temp, left, mid);
    inv_count += _mergeSort(arr, temp, mid+1, right);

    /*Merge the two parts*/
    inv_count += merge(arr, temp, left, mid+1, right);
  }
  return inv_count;
}

/* This funt merges two sorted arrays and returns inversion count in
   the arrays.*/
int merge(int arr[], int temp[], int left, int mid, int right)
{
  int i, j, k;
  int inv_count = 0;

  i = left; /* i is index for left subarray*/
  j = mid;  /* i is index for right subarray*/
  k = left; /* i is index for resultant merged subarray*/
  while ((i <= mid - 1) && (j <= right))
  {
    if (arr[i] <= arr[j])
    {
      temp[k++] = arr[i++];
    }
    else
    {
      temp[k++] = arr[j++];

     /*this is tricky -- see above explanation/diagram for merge()*/
      inv_count = inv_count + (mid - i);
    }
  }

  /* Copy the remaining elements of left subarray
   (if there are any) to temp*/
  while (i <= mid - 1)
    temp[k++] = arr[i++];

  /* Copy the remaining elements of right subarray
   (if there are any) to temp*/
  while (j <= right)
    temp[k++] = arr[j++];

  /*Copy back the merged elements to original array*/
  for (i=left; i <= right; i++)
    arr[i] = temp[i];

  return inv_count;
}

/* Driver progra to test above functions */
int main(int argv, char** args)
{
  int arr[] = {1, 20, 6, 4, 5};
  printf(" Number of inversions are %d \n", mergeSort(arr, 5));
  getchar();
  return 0;
}

这里详细解释： http://www.geeksforgeeks.org/counting-inversions/

使代码反转计数运行得更快

1 个答案: