Worst-case complexity of this function?

Hi,

Could you tell me why the worst-case complexity of the sort() function is O(nlg(n))?

Code:

---

struct s1 {
        long code;
        struct s1 *next;
};

struct s2 {
        long code;
        struct s2 *left;
        struct s2 *right;
};

typedef struct s1 record;
typedef struct s2 node;

...

void sort(record **head) {
        record *p;
        node *root = NULL;
        long t;
       
        if (*head == NULL) {
                return;
        }
        p = *head;
        do {
                add(&root, p->code);        // Worst-case complexity O(lg(n))
                p = p->next;
        } while (p != NULL);        // Will run n times (there are n elements in the list)
        p = *head;
        inOrder(add, &p);        // Adds values to the list in sorted order. Complexity of in-order transversal is O(n), right?
}

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This function is supposed to sort a linked list using a binary search tree. It is an example from one of my school books. Here's how I calculated the worst-case complexity:

n * lg(n) + n => O(n), because n is of higher growth order than n * lg(n)

What am I doing wrong?

searching a linked list is O(N)
searching a already sorted binary tree is O(log N)


QuickSort/BubbleSort/Selection worst case is O(N squared)

HeapSort worst case is O(N log N)

Quote:

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Originally Posted by Ariod

Here's how I calculated the worst-case complexity:

n * lg(n) + n => O(n), because n is of higher growth order than n * lg(n)

What am I doing wrong?

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Are you sure n is of higher growth order than n*lg(n)? I think not.

Quote:

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Originally Posted by hk_mp5kpdw

Are you sure n is of higher growth order than n*lg(n)? I think not.

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Shouldn't it be:

O(1) < O(log(n)) < O(n) < O(nlog(n)) < O(n^2)... for large n's?

Or am I missing something?


And thanks for the input ILoveVectors but I already know that.