Chapter 18, Problem 34E

a.

Explanation of Solution

Representing the numbers in a linked list:

• The numbers can be represented in a linked list by representing each digit in a node.
• Then, the digits in the number 1066 is 1, 0, 6, and 6 which can be represented separately in a linked list as shown:

b.

Explanation of Solution

Addition of two numbers in two different lists and storing the results in another list:

• To add the numbers present in two linked lists, the values present in each node should be added from right to left.
  • The carry should be added with the previous node values...

c.

Explanation of Solution

Algorithm to add two numbers in two different lists and storing the results in another list:

// Set the current pointer of first linked list to point

// the last node of the first linked list.

Set ptr1 to last

// Set the current pointer of second linked list to point

// the last node of the second linked list.

Set ptr2 to last

// Set the value 0 to the variable “remainder”

Set remainder to 0

// Continue the following set of statements until the first // linked list and second linked list becomes empty.

While (ptr1 <> NULL and ptr2 <> NULL)

  // Create a new node.

  Get a node_new

// Add the value of the node where the current pointer

  // of first linked list points with the value of the

// node where the current pointer of second linked

// list points.

// Then by using the MOD function perform the division

// to get the remainder of the additional values and

  // set it to new node.

Set info(node_new) to (info(ptr1) + info(ptr2) + remainder) MOD 10

// Add the value of the node where the current pointer

  // of first linked list points with the value of the

// node where the current pointer of second linked

// list points.

// Then by using the DIV function perform the division // to get the quotient of the additional values and

// set it to new node

Set remainder to (info(ptr1) + info(ptr2) + remainder)DIV 10

// Set the current pointer of first linked list to

  // point the previous node of the current node of the

  // first linked list...