求逆序对个数(合并排序)

求逆序对个数(合并排序)
                 <5,4,2,1,3,6,7,8>------------------------代价是C2*8(C2是常数)
            <5,4,2,1>         <3,6,7,8>-------------------代价是C2*8(C2是常数)
        <5,4>     <2,1>    <3,6>      <7,8>---------------代价是C2*8(C2是常数)
       <5><4>     <2><1>    <3><6>       <7><8>-----------代价是C2*8(C2是常数)
按照上面的方法递归地排序这个数列，对子序列 s1和s2在归并时，若s1[i]>s2[j]（逆序状况），
则逆序数加上s1.length-i,因为s1中i后面的数字对于s2[j]都是逆序的。

上面的算法代价是C2*(8 * (lg8 + 1))= C2*32。
但是，当数据量非常大的时候，比如说n个，那么代价就是C2*(n * (lgn + 1)),也就是说算法的时间复杂度是O(nlgn)
具体实现方法如下(Java)：

public class MergeSorter {private static int counter;public static void main(String[] args) {int number[] = { 4, 3, 2, 1, 0 };compute(number, 0, number.length - 1);for (int i = 0; i < number.length; i++) {System.out.print(number[i] + " ");}System.out.println();System.out.println(counter);}public static void compute(int[] number, int start, int end) {if (start < end) {int mid = (start + end) / 2;compute(number, start, mid);compute(number, mid + 1, end);merge(number, start, mid, end);}}private static void merge(int[] number, int start, int mid, int end) {int merged[] = new int[end - start + 1];int i = start;int j = mid + 1;int k = 0;while (i <= mid && j <= end) {if (number[i] <= number[j]) {merged[k++] = number[i++];} else {counter += mid - i + 1;merged[k++] = number[j++];}}while (i <= mid) {merged[k++] = number[i++];}while (j <= end) {merged[k++] = number[j++];}for (i = start, k = 0; i <= end;) {number[i++] = merged[k++];}}}