如何使用归并排序计算大量输入的反转次数

Posted 2023-03-27

【中文标题】如何使用归并排序计算大量输入的反转次数

【英文标题】：How to count number of inversions for large number of inputs using merge sort

【发布时间】：2020-01-27 08:36:59

【问题描述】：

我能够计算少量输入的反转。我应该进行哪些更改来计算 100000 个输入的反转次数。

#include<iostream>

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);

int mergeSort(int arr[], int array_size)

    int temp[array_size];
    return _mergeSort(arr, temp, 0, array_size - 1);


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 = (left + right) / 2;

        /*Inversion count will be sum of inversions in left-part
         right-part as well as number of inversions while merging */

        inv_count = _mergeSort(arr, temp, left, mid); // counting inversions in the left part
        inv_count += _mergeSort(arr, temp, mid + 1, right); //counting inversions in the right part
        inv_count += merge(arr, temp, left, mid + 1, right);
    
    return inv_count;


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


    int i, j, k;
    int inv_count = 0;

    i = left; // i is the index for left subarray
    j = mid; // j is the index for right subarray
    k = right; // k is the 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++];
            inv_count = inv_count + (mid - i);
        
    

    // copying the remaining parts of the left subarray to temp (if any)
    while (i <= (mid - 1))
        temp[k++] = arr[i++];

    // copying the remaining parts of the left subarray to temp (if any)
    while (j <= right)
        temp[k++] = arr[j++];

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

    return inv_count;


int main()

    int arr[] =  5, 7, 2, 3, 9, 1 ;

    int n = sizeof(arr) / sizeof(arr[0]);

    int ans = mergeSort(arr, n);
    cout << "Number of inversions are" << "\n" << ans;
    return 0;

【问题讨论】：

如果您的基于归并排序的例程正常工作小输入，它也能够处理任何可能大小（亿）的输入

【参考方案1】：

k = 对； // k 是结果合并子数组的索引

这行有问题我猜应该是

k=左;

所以问题出在排序算法上。 合并排序的时间复杂度是 Nlog(N) 而你添加的只是一行

        inv_count = inv_count + (mid - i);

恒定复杂度因此它不会改变算法的整体复杂度。即使输入是100000，计算数组中的反转次数也不会有任何问题。