[剑指offer] 35. 数组中的逆序对

class Solution
{
    int res = 0;
    // 并
    void Merge(vector<int> &target, vector<int> &copy, int left, int right, int mid)
    {
        int i = left;    // 左半部分的索引
        int j = mid + 1; // 右半部分的索引
        int k = 0;         // 被插入数组的索引

        while (i <= mid && j <= right)
        {
            if (target[i] > target[j])
            {
                res += mid - i + 1;
                res%=1000000007;
                copy[k++] = target[j++];
            }
            else
                copy[k++] = target[i++];
        }
        while (i <= mid)
            copy[k++] = target[i++];

        while (j <= right)
            copy[k++] = target[j++];
        k = 0;
        while (left <= right)
        {
            target[left++] = copy[k++];
        }
    }
    // 归
    void MergeSort(vector<int> &target, vector<int> &copy, int left, int right)
    {
        if (left < right)
        {
            int mid = (left + right) / 2;
            MergeSort(target, copy, left, mid);
            MergeSort(target, copy, mid + 1, right);
            Merge(target, copy, left, right, mid);
        }
    }

    void MergeSort(vector<int> &target)
    {
        vector<int> copy(target.size());
        MergeSort(target, copy, 0, target.size() - 1);
    }

  public:
    int InversePairs(vector<int> data)
    {
        if (data.size() == 0)
            return 0;
        MergeSort(data);
        return res;
    }
};