*[codility]ArrayInversionCount

http://codility.com/demo/take-sample-test/arrayinversioncount

求逆序对数，归并排序并记录逆序次数。

// you can also use includes, for example:

// #include <algorithm>

int merge(vector<int> &A, int left, int right) {

    if (left >= right) return 0;

    int mid = left + (right - left) / 2;

    int left_count = merge(A, left, mid);

    int right_count = merge(A, mid + 1, right);

    vector<int> tmp;

    int i = left;

    int j = mid + 1;

    int merge_count = 0;

    while (i <= mid || j <= right) {

        if (i <= mid && j <= right) {

            if (A[i] > A[j]) {

                tmp.push_back(A[i++]);

                merge_count += (right - j + 1);

            }

            else {

                tmp.push_back(A[j++]);

            }

        }

        else if (i <= mid) {

            tmp.push_back(A[i++]);    

        }

        else {

            tmp.push_back(A[j++]);

        }

    }

    for (int k = 0; k < tmp.size(); k++) {

        A[left + k] = tmp[k];

    }

    return (left_count + right_count + merge_count);

}



int solution(const vector<int> &A) {

    // write your code in C++98

    vector<int> B(A);

    return merge(B, 0, B.size() - 1);

}