统计数组中的逆序对的数量

package java_study.JianZhiOffer;

import org.junit.Test;

/**
* Created by ethan on 2015/7/7.
* 剑指offer中No36，直接双重for循环的时间复杂度是O(n2),这样效率太低
* 可以是用归并排序的思路进行排计算逆序对的数目,时间复杂度是O(nlogn)
*/
public class No36数组中的逆序对 {

public int findReversePairNumbers(int[] arr){
if (arr==null) return 0;
if (arr.length==1 || arr.length==0) return 0;
return mergeSort(arr, 0, arr.length-1, new int[arr.length]);
}

public int mergeSort(int[] arr, int start, int end, int[] tmp){
if (start < end){
int mid = (start+end)/2;
int left = mergeSort(arr, start, mid, tmp);
int right = mergeSort(arr, mid+1, end, tmp);
int count = mergeTwoSortedArr(arr, start, mid, end, tmp);
return count+left+right;
}else {
return 0;
}
}

public int mergeTwoSortedArr(int[] arr, int start, int mid, int end, int[] tmp){
int start_1 = start;
int end_1 = mid;
int start_2 = mid+1;
int end_2 = end;
int index = end;
int ans = 0;
while (start_1<=end_1 && start_2<=end_2){
if (arr[end_1]<arr[end_2]){
tmp[index--] = arr[end_2--];
}else {
tmp[index--] = arr[end_1--];
ans += (end_2-start_2+1);
}
}
while (start_1<=end_1){
tmp[index--] = arr[end_1--];
}
while (start_2<=end_2){
tmp[index--] = arr[end_2--];
}
for (int i=start; i<=end; i++){
arr[i] = tmp[i];
}
return ans;
}

@Test
public void test(){
int[] arr = {7, 5, 6, 4};
System.out.println(findReversePairNumbers(arr));
}
}