题解 | #数组中的逆序对#

public class Solution {
    private int res = 0;

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

    private void mergeSort(int[] arr, int left, int right) {
        if (left < right) {
            int mid = left + (right - left) / 2;
            mergeSort(arr, left, mid);
            mergeSort(arr, mid + 1, right);
            merge(arr, left, mid, right);
        }
    }

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