归并排序法求数组中的倒置数量

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如果一对元素（A[i], A[j]）是倒序的，即i < j但是A[i] > A[j],则他们被称为一个倒置。设计o(nlogn)的算法来计算数组中的倒置数量。

利用归并排序实现源码如下：

#include <iostream>#include <cassert>using namespace std;void Merge(int *left, int leftSize, int *right, int rightSize, int *result, int &reverseNum){int *left_end = left + leftSize;int *right_end = right + rightSize;while(left < left_end && right < right_end) {if(*left > *right) {*result++ = *right++;reverseNum += (left_end - left);} else*result++ = *left++;}while(left < left_end)*result++ = *left++;while(right < right_end)*result++ = *right++;}void MergeSort(int *arr, int len, int &reverseNum){if(len == 1)return;int leftSize = len / 2, rightSize = len - len / 2;int *left = new int[leftSize];int *right = new int[rightSize];for(int i = 0; i < leftSize; i++)left[i] = arr[i];for(int i = 0; i < rightSize; i++)right[i] = arr[i + leftSize];MergeSort(arr, leftSize, reverseNum);MergeSort(arr + leftSize, rightSize, reverseNum);Merge(left, leftSize, right, rightSize, arr, reverseNum);delete[] left;delete[] right;}int calReversePair(int *arr, int len){assert(arr);int reverseNum = 0;MergeSort(arr, len, reverseNum);return reverseNum;}int main(){int a[] = {8, 7, 6, 5, 4, 3, 2, 1};cout << calReversePair(a, sizeof(a) / sizeof(a[0])) << endl;return 0;}

运行结果为：28

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