# CSCE 3110 Data Structures & Algorithm Analysis

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CSCE 3110 Data Structures & Algorithm Analysis

Stacks and Queues Implemented with Linked Lists Polynomials Implemented with Linked Lists Remember the array based implementation? Hint: two strategies, one efficient in terms of space, one in terms of running time

Running time? insert, remove traverse, swap How to reverse the elements of a list?

Polynomials Representation typedef struct poly_node *poly_pointer;
int coef; int expon; poly_pointer next; }; poly_pointer a, b, c; coef expon link

Example a null b -3 10 null

Adding Polynomials 3 14 2 8 1 0 a 8 14 -3 10 10 6 b 11 14
2 8 1 0 a -3 10 b 11 14 a->expon == b->expon d 2 8 1 0 a -3 10 b 11 14 -3 10 a->expon < b->expon d

2 8 1 0 a -3 10 b 11 14 -3 10 2 8 d a->expon > b->expon

poly_pointer padd(poly_pointer a, poly_pointer b) { poly_pointer front, rear, temp; int sum; rear =(poly_pointer)malloc(sizeof(poly_node)); if (IS_FULL(rear)) { fprintf(stderr, “The memory is full\n”); exit(1); } front = rear; while (a && b) { switch (COMPARE(a->expon, b->expon)) {

case -1: /* a->expon < b->expon */
attach(b->coef, b->expon, &rear); b= b->next; break; case 0: /* a->expon == b->expon */ sum = a->coef + b->coef; if (sum) attach(sum,a->expon,&rear); a = a->next; b = b->next; case 1: /* a->expon > b->expon */ attach(a->coef, a->expon, &rear); a = a->next; } for (; a; a = a->next) for (; b; b=b->next) rear->next = NULL; temp = front; front = front->next; free(temp); return front;

where m (n) denotes the number of terms in A (B). (2) exponent comparisons extreme case em-1 > fm-1 > em-2 > fm-2 > … > e0 > f0 m+n-1 comparisons (3) creation of new nodes m + n new nodes summary O(m+n)

Attach a Term void attach(float coefficient, int exponent,
poly_pointer *ptr) { /* create a new node attaching to the node pointed to by ptr. ptr is updated to point to this new node. */ poly_pointer temp; temp = (poly_pointer) malloc(sizeof(poly_node)); if (IS_FULL(temp)) { fprintf(stderr, “The memory is full\n”); exit(1); } temp->coef = coefficient; temp->expon = exponent; (*ptr)->next = temp; *ptr = temp;

Other types of lists: Circular lists Doubly linked lists

circular list vs. chain ptr 2 8 1 0 avail ptr temp avail ...

Operations in a circular list
What happens when we insert a node to the front of a circular linked list? X  X  X  a Problem: move down the whole list. A possible solution: X  X  X  a Keep a pointer points to the last node.

Insertion ptr (2) (1) void insertFront (pnode* ptr, pnode node) {
/* insert a node in the list with head (*ptr)->next */ if (IS_EMPTY(*ptr)) { *ptr= node; node->next = node; /* circular link */ } else { node->next = (*ptr)->next; (1) (*ptr)->next = node; (2) X  X  X  ptr (2) (1)

List length int length(pnode ptr) { pnode temp; int count = 0;
if (ptr) { temp = ptr; do { count++; temp = temp->next; } while (temp!=ptr); } return count;

Doubly Linked List Keep a pointer to the next and the previous element in the list typedef struct node *pnode; typedef struct node { char data [4]; pnode next; pnode prev; }

Doubly Linked List Keep a header and trailer pointers (sentinels) with no content header.prev = null; header.next = first element trailer.next = null; trailer.prev = last element Update pointers for every operation performed on the list How to remove an element from the tail of the list ?

Running time? How does this compare to simply linked lists?

Revisit Sparse Matrices
Previous scheme: represent each non-NULL element as a tuple (row, column, value) New scheme: each column (row): a circular linked list with a head node

Nodes in the Sparse Matrix
down col row right entry node value i j aij aij