Given an array nums, we call (i, j) an important reverse pair if i < j and nums[i] > 2*nums[j].

You need to return the number of important reverse pairs in the given array.

Example1:

Input: [1,3,2,3,1]
Output: 2

Example2:

Input: [2,4,3,5,1]
Output: 3

Note:
The length of the given array will not exceed 50,000.
All the numbers in the input array are in the range of 32-bit integer.

思路：
将数组分成左右两个数组，并分别排序，之后对于左边数组的每一个数m，只要找出右边的数组的第一个比m/2大的数即可得出数m的逆序数对个数(因为右边数组是排好序的，第一个比m/2大的数前面的数都可以和m构成逆序数对)。将这些数的逆序数对个数加起来，再加上两个子数组的逆序数对个数即可。

class Solution {
    public int reversePairs(int[] nums) {  
        return mergeSort(nums, 0, nums.length-1);  
    }  
    private int mergeSort(int[] nums, int s, int e){  
        if(s>=e) return 0;   
        int mid = s + (e-s)/2;   
        int cnt = mergeSort(nums, s, mid) + mergeSort(nums, mid+1, e);   
        for(int i = s, j = mid+1; i<=mid; i++){  
            while(j<=e && nums[i]/2.0 > nums[j]) j++;   
            cnt += j-(mid+1);   
        }  
        Arrays.sort(nums, s, e+1);   
        return