Java implements reverse order pair in an array

For two numbers in the array, if the previous number is greater than the latter number, the two numbers form a reverse order pair. For example, in the array {7, 5, 6, 4}, there are a total of 5 reverse order pairs, which are {7, 6}, {7, 5}, {7, 4}, {6, 4}, {5, 4}. Input an array, find the total number P of inverse pairs in this array. And output the result of P modulo 1000000007. , that is, output P%1000000007.

code

Solution one

Violence is simple and inefficient, and will not change the original array

    public static int inversePairs(int[] array) {
        if (array == null || array.length < 2) {
            return 0;
        }
        int count = 0;
        for (int i = 0; i < array.length; i++) {
            for (int j = i + 1; j < array.length; j++) {
                if (array[i] > array[j]) {
                    count++;
                }
            }
        }
        return count % 1000000007;
    }

Solution two

Use the merge sort of the array, which is efficient, but will change the original array

    public static int inversePairs2(int[] array) {
        if (array == null || array.length < 2) {
            return 0;
        }
        int count = mergeSort(array, 0, array.length - 1);
        return count % 1000000007;
    }

    private static int mergeSort(int[] array, int start, int end) {
        if (start >= end) {
            return 0;
        }

        // Find the midpoint of the array, split it into two sub-arrays, and solve it recursively
        int middle = (start + end) / 2;
        int left = mergeSort(array, start, middle);
        int right = mergeSort(array, middle + 1, end);

        // store the merged array
        int[] copy = new int[array.length];
        System.arraycopy(array, start, copy, start, end - start + 1);
        // Start traversing from the end of the two subarrays
        int i = middle;
        int j = end;
        int copyIndex = end;
        // record the number of reverse pairs
        int count = 0;

        while (i >= start && j >= middle + 1) {
            // array is in ascending order
            // If the left array is larger than the right array, put the larger one into the storage array
            // And accumulate the reverse order pairs, it should be ordered, so the i-th element of the left array is larger than the j-th and previous numbers
            if (array[i] > array[j]) {
                copy[copyIndex--] = array[i--];
                count += (j - middle);
            } else {
                copy[copyIndex--] = array[j--];
            }
        }

        // Write the remainder of the subarray to the merged storage array once
        while (i >= start) {
            copy[copyIndex--] = array[i--];
        }
        while (j >= middle + 1) {
            copy[copyIndex--] = array[j--];
        }

        // Write the merge of the two subarrays into the original array
        for (int k = start; k <= end ; k++) {
            array[k] = copy[k];
        }
        return left + right + count;
    }