Question

void merge(vector<int> arr, int l, int m, int r,int &count)
{
    int i, j, k;
    int n1 = m - l + 1;
    int n2 =  r - m;

    /* create temp arrays */
    vector<int> L, R;

    /* Copy data to temp arrays L[] and R[] */
    for (i = 0; i < n1; i++)
        L.push_back(arr[l+i]);
    for (j = 0; j < n2; j++)
        R.push_back(arr[m+1+j]);
     /* Merge the temp arrays back into arr[l..r]*/
    i = 0; // Initial index of first subarray
    j = 0; // Initial index of second subarray
    k = l; // Initial index of merged subarray
    while (i < n1 && j < n2)
    {
        if (L[i] <= R[j])
        {
            arr[k] = L[i];
            i++;
        }
        else
        {
            arr[k] = R[j];
            count+=(n1-i);
            j++;
        }
        k++;
    }

    /* Copy the remaining elements of L[], if there
       are any */
    while (i < n1)
    {
        arr[k] = L[i];
        i++;
        k++;
    }

    /* Copy the remaining elements of R[], if there
       are any */
    while (j < n2)
    {
        arr[k] = R[j];
        j++;
        k++;
    }
}

/* l is for left index and r is right index of the
   sub-array of arr to be sorted */
void mergeSort(vector<int> arr, int l, int r,int &count)
{
    if (l < r)
    {
        // Same as (l+r)/2, but avoids overflow for
        // large l and h
        int m = l+(r-l)/2;

        // Sort first and second halves
        mergeSort(arr, l, m,count);
        mergeSort(arr, m+1, r,count);

        merge(arr, l, m, r,count);
    }
}

int Solution::countInversions(vector<int> &A) {
    int count = 0;
    mergeSort(A,0,A.size()-1,count);
    return count;
}

这是我的计数反转次数的代码。我也尽早解决了这个问题，所以我对实现的逻辑很有信心，我不知道为什么它在面试“比特测试用例”时没有通过某些测试用例。任何帮助将不胜感激！谢谢。

您的提交因以下输入而失败：

A : [ 84, 2, 37, 3, 67, 82, 19, 97, 91, 63, 27, 6, 13, 90, 63, 89, 100, 60, 47, 96, 54, 26, 64, 50, 71, 16, 6, 40, 84, 93, 67, 85, 16, 22, 60 ]

您的函数返回了以下内容：

期望的返回值：

Answer 1

您正在计算递归的每个级别的反转；这个问题（显然）只对输入数组中的反转感兴趣。

如果将您对反转的定义应用于输入数组，则会得到预期的返回值。

倒数计数答案错误？

1 个答案: