Reverse Pairs

For an array A, if i < j, and A [i] > A [j], called (A [i], A [j]) is a reverse pair.
return total of reverse pairs in A.

 Example

Given A = [2, 4, 1, 3, 5] , (2, 1), (4, 1), (4, 3) are reverse pairs. return 3

分析：

如果用两个for loop,非常简单。但要降低复杂度，做法非常变态.简单来说，原理就是利用merge sort的方法，把左subarray 和右subarray 合并（假设它们都已排好序）。然后比较当前左subarray的第一个值是否比右边subarray第一个值大两倍，如果是，那么很明显左subarray所有的都比那个值大两倍。

 class Solution {
     public int reversePairs(int[] nums) {
         return mergeSort(nums, , nums.length - );
     }     public int mergeSort(int[] nums, int start, int end) {
         if (start >= end) return ;
         int total =  ;
         int mid = start + (end - start) / ;
         total += mergeSort(nums, start, mid);
         total += mergeSort(nums, mid + , end);
         total += merge(nums, start, mid, end);
         return total;
     }     public int merge(int[] nums, int start, int mid, int end) {
         int[] temp = new int[end - start + ];
         int p1 = start, p2 = mid + ;
         int index = ,count = ;         while (p1 <= mid && p2 <= end) {
             if ((long) nums[p1] >  * ((long) nums[p2])) {
                 count += mid - p1 + ;
                 p2++;
             } else {
                 p1++;
             }
         }
         p1 = start;
         p2 = mid + ;
         while (p1 <= mid && p2 <= end) {
             if (nums[p1] < nums[p2]) {
                 temp[index++] = nums[p1++];
             } else {
                 temp[index++] = nums[p2++];
             }
         }         while(p1 <= mid) {
             temp[index++] = nums[p1++];
         }
         while(p2 <= end) {
             temp[index++] = nums[p2++];
         }
         System.arraycopy(temp, , nums, start, end - start + );
         return count;
     }
 }