O(nlogn)算法会花费多长时间？

假设有一个包含 n 个元素的数组，使用 O(n log n) 算法对其排序。代码示例如下：

#include <stdio.h>

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[n1], R[n2];

    /* 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++;
    }
}

void merge_sort(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
        merge_sort(arr, l, m);
        merge_sort(arr, m + 1, r);

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

int main() {
    int arr[] = {12, 11, 13, 5, 6, 7};

    int n = sizeof(arr) / sizeof(arr[0]);

    merge_sort(arr, 0, n - 1);

    printf("Sorted array is \n");
    for (int i = 0; i < n; i++)
        printf("%d ", arr[i]);
    printf("\n");

    return 0;
}

该算法的时间复杂度为 O(n log n)，