Question

this code is working when array size is 75.000 but I get stack overflow error when it's 100.000 what can I do to fix it? Program needs to calculate the elapsed time and average time for both algorithms and compare them

This is where error begins

 //_______________________________
    // MERGE SORT
    void merge(int arr[], int l, int m, int r)
    {
        int i, j, k;
        int n1 = m - l + 1;
        int n2 = r - m;

        /* create temp arrays */
    int L[100000], R[100000];

    /* Copy data to temp arrays L[] and R[] */
    for (i = 0; i < n1; i++)
        L[i] = arr[l + i];
    for (j = 0; j < n2; j++)
        R[j] = 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];
            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++;
    }
}

some details for posting!..

/* l is for left index and r is right index of the
sub-array of arr to be sorted */
void mergeSort(int arr[], int l, int r)
{
    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);
        mergeSort(arr, m + 1, r);

        merge(arr, l, m, r);
    }
}
// END MERGE SORT
//_______________________________

Insertion part does not give any error on 100.000 but merge part does

Answer 1

堆栈大小非常小，因此请尝试在堆中分配这两个大型数组

int *L = new int[100000];
int *R = new int[100000];

... merge stuff ...

delete [] L;
delete [] R;

由于L的大小为n1且R的大小为n2，因此您可以

int *L = new int[n1];
int *R = new int[n2];

merge sort stack overflow in c++

1 个答案: