【InversionCount 逆序对数 + MergeSort】

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Definition of Inversion: Let (A[0], A[1] ... A[n], n <= 50) be a sequence of n numbers. If i < j and A[i] > A[j], then the pair (i, j) is called inversion of A.

Example:

Count(Inversion({3, 1, 2})) = Count({3, 1}, {3, 2}) = 2

思路，如果用brute force，则O（n^2），借用合并排序里面的合并步骤里的思路

import java.util.Arrays;public class MergeSort {static int InversionCount  = 0;public static void main(String[] args) {int[] array = {3,1,2,5,4,7,6};MergeSort(array, 0, array.length-1);System.out.println(InversionCount);System.out.println(Arrays.toString(array));}public static void MergeSort(int[] array, int lhs, int rhs) {if (lhs < rhs) {int mid = lhs + ((rhs - lhs)>>1);MergeSort(array, lhs, mid);MergeSort(array, mid+1, rhs);Merge(array, lhs, mid, rhs);}}public static void Merge(int[] array, int lhs, int mid, int rhs) {int[] tmp = new int[rhs-lhs+1];int i = lhs, j = mid+1;int k = 0;while(i <= mid && j <= rhs){if (array[i] > array[j]) {InversionCount += mid-i+1;tmp[k++] = array[j++];}else {tmp[k++] = array[i++];}}while(i <= mid){tmp[k++] = array[i++];}while(j <= rhs){tmp[k++] = array[j++];}for (i = 0; i < k; i++) {array[i+lhs] = tmp[i];}tmp = null;}}

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