【InversionCount 逆序对数 + MergeSort】

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.

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

思路，假设用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;
	}

}