【剑指Offer】数组中的逆序对

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题目

在数组中的两个数字如果前面一个数字大于后面的数字，则这两个数字组成一个逆序对，输入一个数组，求出这个数组中的逆序对的总数。

实现

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

    int[] aux = new int[array.length];

    for (int i = 0; i < array.length; i++)
        aux[i] = array[i];

    return sortAndCount(array, aux, 0, array.length - 1);
}

private int sortAndCount(int[] array, int[] aux, int start, int end){
    if (start == end)
        return 0;

    int mid = (start + end) / 2;

    int leftCount = sortAndCount(array, aux, start, mid);
    int rightCount = sortAndCount(array, aux, mid + 1, end);
    int splitCount = countSplitInv(array, aux, start, mid, end);

    return (leftCount + rightCount + splitCount) % 1000000007;
}

private int countSplitInv(int[] array, int[] aux, int start, int mid, int end) {
    int count = 0;
    int j = start;
    int k = mid + 1;

    for (int i = start; i <= end; i++) {
        if (j > mid)
            array[i] = aux[k++];
        else if (k > end)
            array[i] = aux[j++];
        else if (aux[j] < aux[k])
            array[i] = aux[j++];
        else {
            count += mid - j + 1;
            array[i] = aux[k++];

            if (count > 1000000007)
                count %= 1000000007;
        }
    }

    for (int i = start; i <= end; i++)
        aux[i] = array[i];

    return count;
}