有一组数，对于其中任意两个数组，若前面一个大于后面一个数字，则这两个数字组成一个逆序对。请设计一个高效的算法，计算给定数组中的逆序对个数。 给定一个int数组A和它的大小n，请返回A中的逆序对个数。保证n小于等于5000。 测试样例： [1,2,3,4,5,6,7,0],8 返回：7
package com.mico.alex; import java.util.Scanner; public class GetReverseNum { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub Scanner scanner = new Scanner(System.in); while (scanner.hasNext()) { int n = scanner.nextInt(); int[] A = new int[n]; for (int i = 0; i < n; i++) { A[i] = scanner.nextInt(); } System.out.println(count(A, n)); } } public static int count(int[] A, int n) { return mergeSort(A, 0, n - 1); } public static int merge(int[] A, int low, int mid, int high) { int[] B = new int[high - low + 1]; int i = low; int j = mid + 1; int index = 0; int reverseNum = 0; while (i <= mid && j <= high) { if (A[i] <= A[j]) { B[index++] = A[i]; i++; } else { B[index++] = A[j]; j++; reverseNum += mid - i + 1; } } while (i <= mid) { B[index++] = A[i++]; } while (j <= high) { B[index++] = A[j++]; } for (int k = 0; k < B.length; k++) { A[low + k] = B[k]; } return reverseNum; } public static int mergeSort(int[] A, int start, int high) { if (start >= high) { return 0; } int mid = (start + high) / 2; int count1 = mergeSort(A, start, mid); int count2 = mergeSort(A, mid + 1, high); return count1 + count2 + merge(A, start, mid, high); } }