class Solution {

     public int reversePairs(int[] nums) {

         //给定数组，求符合条件的逆序对

         //思路:同剑指offer的类似，但是每次合并需要从头开始合并，同时排序

         if(nums==null||nums.length==0) return 0;

         return reversePairsCore(nums,0,nums.length-1);

     }

     public int reversePairsCore(int [] nums,int start,int end){

         if(start>=end){

             return 0;

         }else{

             int mid=start+(end-start)/2;

             int left=reversePairsCore(nums,start,mid);

             int right=reversePairsCore(nums,mid+1,end);

             int [] copy=new int[end-start+1];

             int i=start,t=start;

             int j=mid+1;

             int curIndex=0;

             int count=0;

             for(;j<=end;j++,curIndex++){

                 //查找满足条件的

                 while(i<=mid&&nums[i]<=2*(long)nums[j]){

                     i++;

                 }

                 //排序

                 while(t<=mid&&nums[t]<nums[j]){

                     copy[curIndex++]=nums[t++];

                 }

                 //将j放入

                 copy[curIndex]=nums[j];

                 count+=mid-i+1;

             }

             //防止还有后面大的元素

             while(t<=mid){

                 copy[curIndex++]=nums[t++];

             }

             //将数组赋值给nums

             System.arraycopy(copy,0,nums,start,end-start+1);

             return left+count+right;

         }

     }

 }