Question

Below are the node_struct, link, list_struct and list as defined in class and in hw3: typedef struct node_struct * link; struct node_struct { int item; link next; };

typedef struct list_struct * list; struct list_struct { link first; int length; };

a) (15 pts) Write a function that swaps the first and last node in a list. If there are not enough nodes, it does not do anything to the list. You have to update the links yourself. DO NOT call any insert or remove functions to do any of the work for you.
void swapFirstLast(list L) {
b) (8 pts) Make a drawing of a list with 5 (five) nodes and show how the links are updated as done in class. Label each arrow update with the line number that did it.
c) (3 pts) What is the Θ complexity of your function? Justify your answer. (You can show to the side of the code)
d) (4 pts) If your function fails or crashes for certain inputs, give those inputs and clearly indicate the line with the problem (use line numbers or write the test cases as a comment on that line of code).

Answer 1

void swapFirstLast(list L) {
    // No swaps required on less than length 2.
    if(L == NULL || L->length <= 1) {
        return;
    }

    // special case for length of 2 nodes.
    if(L->length == 2) {
        link ourfirst = L->first;
        link oursecond = ourfirst->next;

        ourfirst->next = NULL;
        oursecond->next = ourfirst;

        // make the last node, as first
        L->first = oursecond;
    } else {
        link ourfirst = L->first;
        link ourlastPrev = L->first;
        link ourlast = L->first->next;

        while(ourlast->next != NULL) {
            ourlastPrev = ourlast;
            ourlast = ourlast->next;
        }
        L->first = ourlastPrev->next;
        L->first->next = ourfirst->next;
        ourlastPrev->next = ourfirst;       
        ourfirst->next = NULL;
    }
}

Taking directly case 3, with length more than 2 nodes, as other cases seem trivial compared to this..

So on start of line 18, the list looks like below:

From line 18 to line 25, we are trying to find the first node, last and seconlast node.. After line25, the markers will be adjusted as below:

After line 27:

After line 28:

After line 29:

After line 30:

The complexity is O(N) for N nodes, as we need to find the last node to swap. Also, code does not crash for any input.