Question

以下是计算数组中的反转的代码。我对此部分有疑问：

     inv_count  = _mergeSort(arr, temp, left, mid);--i

    inv_count += _mergeSort(arr, temp, mid+1, right);--ii

     inv_count += merge(arr, temp, left, mid+1, right);--iii

在反转计数中，总反转将等于i + ii + iii，但是我无法理解＆＃34; i和ii的inv_count甚至得到一个值，它们被递归调用并填充到函数堆栈但是没有赋值给inv_count 对于i和ii，虽然在iii中，inv_count使用invcount = inv_count + mid-i获取值;

   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;
}

Answer 1

这是反转计数的工作代码.-- i --ii --iii与反转计数无关。

合并排序期间的反转总数等于从排序左侧部分和排序右侧部分以及合并左侧部分和右侧部分获得的反转之和。如果您不清楚代码，请进行注释。

#include <iostream>
#include <stdio.h>
#include <limits.h>
#include <algorithm>
using namespace std;
const int size = 1000000;
long long int array[size];
long long int merge(long long int a[], long long int beg, long long int mid,
    long long int end) {
    long long int inverse = 0;
    long long int lsize = (mid - beg) + 1;
    long long int rsize = (end - mid);
    long long int left[lsize + 1];
    long long int right[rsize + 1];
    long long int i;
    long long int j = beg;
    for (i = 0; i < lsize; ++i, ++j) {
        left[i] = a[j];
    }
    j = mid + 1;
    for (i = 0; i < rsize; ++i, ++j) {
        right[i] = a[j];
    }
    left[lsize] = LONG_LONG_MAX;
    right[rsize] = LONG_LONG_MAX;
    j = 0;
    i = 0;
    for (int k = beg; k <= end; ++k) {
       if (left[i] <= right[j]) {
            a[k] = left[i];
            ++i;
       } else {
            a[k] = right[j];
            inverse += (lsize - i);
            ++j;
      }
    }
   return inverse;
}

long long int merge_sort(long long int iArray[], long long int beg,
        long long int end) {
     if (beg < end) {
        long long int mid;
        long long int left = 0;
        long long int right = 0;
        long long int total = 0;
        mid = (beg + end) / 2;
        left = merge_sort(iArray, beg, mid);
        right = merge_sort(iArray, mid + 1, end);
        total = merge(iArray, beg, mid, end);
        return left + right + total;
     } else {
         return 0;
     }
}

合并排序中的反转计数值

1 个答案: