【InversionCount 逆序对数 + MergeSort】

Definition of Inversion: Let (A[0], A[1] ... A
, 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;
}

}