题目链接：. - 力扣（LeetCode）. - 备战技术面试？力扣提供海量技术面试资源，帮助你高效提升编程技能,轻松拿下世界 IT 名企 Dream Offer。https://leetcode.cn/problems/shu-zu-zhong-de-ni-xu-dui-lcof/submissions/571702329/

代码：

class Solution {
    
    private int[] tmp;

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

    // 归并排序函数
    private int mergeSort(int[] record, int left, int right) {
        if (left >= right) return 0; // 如果只有一个元素或没有元素，不存在逆序对
        int mid = left + (right - left) / 2;
        // 递归分治，分别计算左边和右边的逆序对
        int count = mergeSort(record, left, mid) + mergeSort(record, mid + 1, right);
        // 统计跨子数组的逆序对
        int i = left, j = mid + 1, k = 0;
        while (i <= mid && j <= right) {
            if (record[i] <= record[j]) {
                tmp[k++] = record[i++];
            } else {
                // record[i] > record[j]，因此 (i, j) 是逆序对，并且 i 到 mid 都与 j 构成逆序对
                tmp[k++] = record[j++];
                count += mid - i + 1;
            }
        }
        // 处理剩余的元素
        while (i <= mid) tmp[k++] = record[i++];
        while (j <= right) tmp[k++] = record[j++];
        // 将排好序的部分拷贝回原数组
        for (int l = 0; l < k; l++) {
            record[left + l] = tmp[l];
        }
        return count;
    }
    
    public static void main(String[] args) {
        Solution solution = new Solution();
        int[] record = {9, 7, 5, 4, 6};
        System.out.println(solution.reversePairs(record));  // 输出：8
    }
}