翻转对

/*
题目思路:
我们仍然需要找到数组中的所有重要翻转对(i, j)，满足 i < j 且 nums[i] > 2 * nums[j]。
虽然Java Stream API不太适合进行复杂的索引操作和修改，但我们可以使用Stream API来简化某些部分。
此解决方案主要展示如何结合使用Stream API和传统循环。
*/
import java.util.stream.*;
public class SolutionStream {
    public int reversePairs(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        return mergeSort(nums, 0, nums.length - 1);
    }
 
    private int mergeSort(int[] nums, int left, int right) {
        if (left >= right) return 0;
        int mid = left + (right - left) / 2;
        int count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);
        count += (int) IntStream.range(left, mid + 1).boxed().mapToLong(i -> {
            return IntStream.range(mid + 1, right + 1).filter(j -> nums[i] > 2L * nums[j]).count();
        }).sum();
        merge(nums, left, mid, right);
        return count;
    }
 
    private void merge(int[] nums, int left, int mid, int right) {
        int[] temp = new int[right - left + 1];
        int i = left, j = mid + 1, k = 0;
        while (i <= mid && j <= right) {
            if (nums[i] <= nums[j]) temp[k++] = nums[i++];
            else temp[k++] = nums[j++];
        }
        while (i <= mid) temp[k++] = nums[i++];
        while (j <= right) temp[k++] = nums[j++];
        System.arraycopy(temp, 0, nums, left, temp.length);
    }
}