题解 | #数组中的逆序对#

public class Solution {
    public int InversePairs(int [] array) {
        if (array == null || array.length < 1) {
            return 0;
        }
        int[] res = new int[1];
        mergeSort(array, 0, array.length - 1, res);
        return res[0];
    }

    private void mergeSort(int[] arr, int l, int r, int[] res) {
        if (l >= r) {
            return;
        }
        int mid = l + (r - l) / 2;
        mergeSort(arr, l, mid, res);
        mergeSort(arr, mid + 1, r, res);
        merge(arr, l, mid, r, res);
    }

    private void merge(int[] arr, int l, int mid, int r, int[] res) {
        int[] tmp = new int[r - l + 1];
        int i = l;
        int j = mid + 1;
        int k = 0;
        while (i <= mid && j <= r) {
            if (arr[i] > arr[j]) {
                tmp[k++] = arr[j++];
                res[0] += (mid - i + 1);
                res[0] %= 1000000007;
            } else {
                tmp[k++] = arr[i++];
            }
        }

        while (i <= mid) {
            tmp[k++] = arr[i++];
        }
        while (j <= r) {
            tmp[k++] = arr[j++];
        }
        for (k = 0, i = l; i <= r; i++, k++) {
            arr[i] = tmp[k];
        }
    } 
}