本文概述

• C ++
• C
• Java
• Python3
• C#
• 的PHP
• C ++
• C
• Java
• Python3
• C#

数组的反转计数指示了数组距离被排序有多远(或多近)。如果数组已经排序，则反转计数为0，但如果数组是以相反的顺序排序，则反转计数为最大值。
形式上讲, 如果a [i]> a [j]且i < j, 则两个元素a [i]和a [j]构成一个求逆
【算法（计算数组中的反转(逆序)S1（使用合并排序））】例子：

Input: arr[] = {8, 4, 2, 1} Output: 6Explanation: Given array has six inversions: (8, 4), (4, 2), (8, 2), (8, 1), (4, 1), (2, 1).Input: arr[] = {3, 1, 2} Output: 2Explanation: Given array has two inversions: (3, 1), (3, 2)

方法1(简单)   

• 方法：遍历数组, 并为每个索引找到数组右侧较小元素的数量。这可以使用嵌套循环来完成。对数组中所有索引的计数求和, 然后打印总和。
• 算法： 
  1. 从头到尾遍历数组
  2. 对于每个元素, 使用另一个循环查找小于当前数量的元素数, 直到该索引为止。
  3. 总结每个索引的反转计数。
  4. 打印反转计数。
• 实现 

C ++

// C++ program to Count Inversions // in an array #include < bits/stdc++.h> using namespace std; int getInvCount( int arr[], int n) { int inv_count = 0; for ( int i = 0; i < n - 1; i++) for ( int j = i + 1; j < n; j++) if (arr[i] > arr[j]) inv_count++; return inv_count; } // Driver Code int main() { int arr[] = { 1, 20, 6, 4, 5 }; int n = sizeof (arr) / sizeof (arr[0]); cout < < " Number of inversions are " < < getInvCount(arr, n); return 0; } // This code is contributed // by Akanksha Rai

C

// C program to Count // Inversions in an array #include < stdio.h> #include < stdlib.h> int getInvCount( int arr[], int n) { int inv_count = 0; for ( int i = 0; i < n - 1; i++) for ( int j = i + 1; j < n; j++) if (arr[i] > arr[j]) inv_count++; return inv_count; } /* Driver program to test above functions */ int main() { int arr[] = { 1, 20, 6, 4, 5 }; int n = sizeof (arr) / sizeof (arr[0]); printf ( " Number of inversions are %d \n" , getInvCount(arr, n)); return 0; }

Java

// Java program to count // inversions in an array class Test { static int arr[] = new int [] { 1 , 20 , 6 , 4 , 5 }; static int getInvCount( int n) { int inv_count = 0 ; for ( int i = 0 ; i < n - 1 ; i++) for ( int j = i + 1 ; j < n; j++) if (arr[i] > arr[j]) inv_count++; return inv_count; } // Driver method to test the above function public static void main(String[] args) { System.out.println( "Number of inversions are " + getInvCount(arr.length)); } }

Python3

# Python3 program to count # inversions in an array def getInvCount(arr, n): inv_count = 0 for i in range (n): for j in range (i + 1 , n): if (arr[i] > arr[j]): inv_count + = 1 return inv_count # Driver Code arr = [ 1 , 20 , 6 , 4 , 5 ] n = len (arr) print ( "Number of inversions are" , getInvCount(arr, n)) # This code is contributed by Smitha Dinesh Semwal

C#

// C# program to count inversions // in an array using System; using System.Collections.Generic; class GFG { static int [] arr = new int [] { 1, 20, 6, 4, 5 }; static int getInvCount( int n) { int inv_count = 0; for ( int i = 0; i < n - 1; i++) for ( int j = i + 1; j < n; j++) if (arr[i] > arr[j]) inv_count++; return inv_count; } // Driver code public static void Main() { Console.WriteLine( "Number of " + "inversions are " + getInvCount(arr.Length)); } } // This code is contributed by Sam007

的PHP

< ?php // PHP program to Count Inversions // in an array function getInvCount(& $arr , $n ) { $inv_count = 0; for ( $i = 0; $i < $n - 1; $i ++) for ( $j = $i + 1; $j < $n ; $j ++) if ( $arr [ $i ] > $arr [ $j ]) $inv_count ++; return $inv_count ; } // Driver Code $arr = array (1, 20, 6, 4, 5 ); $n = sizeof( $arr ); echo "Number of inversions are " , getInvCount( $arr , $n ); // This code is contributed by ita_c ?>

• 输出如下： 

Number of inversions are 5

• 复杂度分析： 
  • 时间复杂度：O(n ^ 2), 需要两个嵌套循环才能从头到尾遍历数组, 因此时间复杂度为O(n ^ 2)
  • 空间复杂:O(1), 不需要额外的空间。

方法2(增强合并排序) 

• 方法： 
  假设数组左半边和右半边的反转次数(分别为inv1和inv2), 在Inv1 + Inv2中没有考虑哪些反转？答案是–在合并步骤中需要计算的反转数。因此, 要获得许多反转, 需要在左子数组, 右子数组和merge()中添加许多反转。
   

文章图片

• 怎么获得的merge()中的反转次数？ 
  在合并过程中, 让我用于索引左子数组, 让j用于右子数组。在merge()的任何步骤中, 如果a [i]大于a [j], 则存在(mid – i)个反转。因为左子数组和右子数组已排序, 所以左子数组中的所有其余元素(a [i + 1], a [i + 2]…a [mid])将大于a [j]

文章图片

• 完整图片： 

文章图片

• 算法： 
  1. 这个想法类似于合并排序, 将数组分成两个相等或几乎相等的两半, 直到达到基本情况为止。
  2. 创建一个函数合并, 计算合并数组的两半时的反转次数, 创建两个索引i和j, i是上半部分的索引, j是下半部分的索引。如果a [i]大于a [j], 则存在(mid – i)个反演。因为左子数组和右子数组已排序, 所以左子数组中的所有其余元素(a [i + 1], a [i + 2]…a [mid])将大于a [j]。
  3. 创建一个递归函数, 将数组分成两半, 然后求和前一半的求反数??, 再求后一半的求反数??, 然后将二者合并, 求出答案。
  4. 递归的基本情况是在给定的一半中只有一个元素。
  5. 打印答案
• 实现 

C ++

// C++ program to Count // Inversions in an array // using Merge Sort #include < bits/stdc++.h> using namespace std; 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[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; /* j is index for right subarray*/ k = left; /* k 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 code int main() { int arr[] = { 1, 20, 6, 4, 5 }; int n = sizeof (arr) / sizeof (arr[0]); int ans = mergeSort(arr, n); cout < < " Number of inversions are " < < ans; return 0; } // This is code is contributed by rathbhupendra

C

// C program to Count // Inversions in an array // using Merge Sort #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 the 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; /* j is index for right subarray*/ k = left; /* k 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 program 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; }

Java

// Java implementation of the approach import java.util.Arrays; public class GFG { // Function to count the number of inversions // during the merge process private static int mergeAndCount( int [] arr, int l, int m, int r) { // Left subarray int [] left = Arrays.copyOfRange(arr, l, m + 1 ); // Right subarray int [] right = Arrays.copyOfRange(arr, m + 1 , r + 1 ); int i = 0 , j = 0 , k = l, swaps = 0 ; while (i < left.length & & j < right.length) { if (left[i] < = right[j]) arr[k++] = left[i++]; else { arr[k++] = right[j++]; swaps += (m + 1 ) - (l + i); } } return swaps; } // Merge sort function private static int mergeSortAndCount( int [] arr, int l, int r) { // Keeps track of the inversion count at a // particular node of the recursion tree int count = 0 ; if (l < r) { int m = (l + r) / 2 ; // Total inversion count = left subarray count // + right subarray count + merge count // Left subarray count count += mergeSortAndCount(arr, l, m); // Right subarray count count += mergeSortAndCount(arr, m + 1 , r); // Merge count count += mergeAndCount(arr, l, m, r); } return count; } // Driver code public static void main(String[] args) { int [] arr = { 1 , 20 , 6 , 4 , 5 }; System.out.println(mergeSortAndCount(arr, 0 , arr.length - 1 )); } } // This code is contributed by Pradip Basak

Python3

# Python 3 program to count inversions in an array # Function to Use Inversion Count def mergeSort(arr, n): # A temp_arr is created to store # sorted array in merge function temp_arr = [ 0 ] * n return _mergeSort(arr, temp_arr, 0 , n - 1 ) # This Function will use MergeSort to count inversions def _mergeSort(arr, temp_arr, left, right): # A variable inv_count is used to store # inversion counts in each recursive call inv_count = 0 # We will make a recursive call if and only if # we have more than one elements if left < right: # mid is calculated to divide the array into two subarrays # Floor division is must in case of python mid = (left + right) / / 2 # It will calculate inversion # counts in the left subarray inv_count + = _mergeSort(arr, temp_arr, left, mid) # It will calculate inversion # counts in right subarray inv_count + = _mergeSort(arr, temp_arr, mid + 1 , right) # It will merge two subarrays in # a sorted subarray inv_count + = merge(arr, temp_arr, left, mid, right) return inv_count # This function will merge two subarrays # in a single sorted subarray def merge(arr, temp_arr, left, mid, right): i = left# Starting index of left subarray j = mid + 1 # Starting index of right subarray k = left# Starting index of to be sorted subarray inv_count = 0 # Conditions are checked to make sure that # i and j don't exceed their # subarray limits. while i < = mid and j < = right: # There will be no inversion if arr[i] < = arr[j] if arr[i] < = arr[j]: temp_arr[k] = arr[i] k + = 1 i + = 1 else : # Inversion will occur. temp_arr[k] = arr[j] inv_count + = (mid - i + 1 ) k + = 1 j + = 1 # Copy the remaining elements of left # subarray into temporary array while i < = mid: temp_arr[k] = arr[i] k + = 1 i + = 1 # Copy the remaining elements of right # subarray into temporary array while j < = right: temp_arr[k] = arr[j] k + = 1 j + = 1 # Copy the sorted subarray into Original array for loop_var in range (left, right + 1 ): arr[loop_var] = temp_arr[loop_var]return inv_count # Driver Code # Given array is arr = [ 1 , 20 , 6 , 4 , 5 ] n = len (arr) result = mergeSort(arr, n) print ( "Number of inversions are" , result) # This code is contributed by ankush_953

C#

// C# implementation of counting the // inversion using merge sort using System; public class Test { /* This method sorts the input array and returns the number of inversions in the array */ static int mergeSort( int [] arr, int array_size) { int [] temp = new int [array_size]; return _mergeSort(arr, temp, 0, array_size - 1); } /* An auxiliary recursive method that sorts the input array and returns the number of inversions in the array. */ static 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 the 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 method merges two sorted arrays and returns inversion count in the arrays.*/ static 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; /* j is index for right subarray*/ k = left; /* k 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 method to test the above function public static void Main() { int [] arr = new int [] { 1, 20, 6, 4, 5 }; Console.Write( "Number of inversions are " + mergeSort(arr, 5)); } } // This code is contributed by Rajput-Ji

• 输出如下： 

Number of inversions are 5

• 复杂度分析： 
  • 时间复杂度：O(n log n), 使用的算法是分治法, 因此在每个级别中需要一个完整的数组遍历, 并且存在log n个级别, 因此时间复杂度为O(n log n)。
  • 空间复杂:O(n), 临时数组。

请注意, 上面的代码修改(或排序)了输入数组。如果我们只想计算倒数, 那么我们需要创建一个原始数组的副本, 并在副本上调用mergeSort()。
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使用C++ STL中的Set计数反转
计算数组中的反转第三组(使用BIT)
参考文献：
http://www.cs.umd.edu/class/fall2009/cmsc451/lectures/Lec08-inversions.pdf
http://www.cp.eng.chula.ac.th/~piak/teaching/algo/algo2008/count-inv.htm
如果你在上述程序/算法或其他解决相同问题的方法中发现任何错误, 请发表评论。

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