左程云 逆序对问题

//方法：归并排序，等同于小和问题。升序排序，只要左大于右，左之后所有都大于右，直接加上。若遇到相等问题，先拷贝左
class Solution {//先写出一个标准的归并排序，再对其进行改造。
    public int reversePairs(int[] nums) {
        if(nums == null || nums.length < 2){
            return 0;
        }
        //计算mid，左归并，右归并，左右合并。
        return mergeSort(nums, 0, nums.length-1);
    }

    public int mergeSort(int[] nums, int l, int r){
        if(l == r){
            return 0;
        }
        int mid = l + ((r - l) >> 1);
        return mergeSort(nums, l, mid) + mergeSort(nums, mid+1, r) + merge(nums, r, mid, l);
    }

    public int merge(int[] nums, int r, int mid, int l){
        int[] help = new int[r - l + 1];
        int i = 0;
        int p1 = l;
        int p2 = mid + 1;
        int cnt =0;
        while(p1 <= mid  && p2 <= r){
            if(nums[p1] <= nums[p2]){
                help[i++] = nums[p1++];
            }else if(nums[p1] == nums[p2]){
                help[i++] = nums[p2++];
            }else{
                cnt += mid - p1 + 1;
                help[i++] = nums[p2++];
            }
        }

        while(p1 <= mid){
            help[i++] = nums[p1++];
        }
        while(p2 <= r){
            help[i++] = nums[p2++];
        }
        for(i = 0; i < help.length; i++){
            nums[l+i] = help[i];
        }
        return cnt;
    }
}