分治逆序对计数java实现

以下是基于Java语言的分治逆序对计数算法实现：

public class ReversePairs {
    public static int reversePairs(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int[] temp = new int[nums.length];
        return mergeSort(nums, 0, nums.length - 1, temp);
    }

    private static int mergeSort(int[] nums, int left, int right, int[] temp) {
        if (left >= right) {
            return 0;
        }
        int mid = left + (right - left) / 2;
        int invLeft = mergeSort(nums, left, mid, temp);
        int invRight = mergeSort(nums, mid + 1, right, temp);
        int invMerge = merge(nums, left, mid, right, temp);
        return invLeft + invRight + invMerge;
    }

    private static int merge(int[] nums, int left, int mid, int right, int[] temp) {
        int i = left, j = mid + 1, k = left;
        int invCount = 0;
        while (i <= mid && j <= right) {
            if (nums[i] <= nums[j]) {
                temp[k++] = nums[i++];
            } else {
                temp[k++] = nums[j++];
                invCount += mid - i + 1;
            }
        }
        while (i <= mid) {
            temp[k++] = nums[i++];
        }
        while (j <= right) {
            temp[k++] = nums[j++];
        }
        for (i = left; i <= right; i++) {
            nums[i] = temp[i];
        }
        return invCount;
    }
}

其中，reversePairs函数是算法的入口，用于对数组nums进行逆序对计数，返回总的逆序对数量。mergeSort函数是递归函数，用于将原数组分成左右两个子数组进行处理；merge函数用于将左右两个子数组合并成一个有序的数组，并计算左右两个子数组之间的逆序对数量。整个算法的时间复杂度为O(nlogn)。