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:

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Answer 1

Expert Solution & Answer

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...

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See solution

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Answer 2

Expert Solution & Answer

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...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Expert Solution & Answer

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...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

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...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

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...

Answer 8

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:

Answer 9

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:

Answer 10

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...

Answer 11

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...

Answer 12

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...

Answer 13

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...

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

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Check out a sample textbook solution

See solution

Answer 17

Ch. 18 - Prob. 1E Ch. 18 - Prob. 2E Ch. 18 - Prob. 3E Ch. 18 - Prob. 4E Ch. 18 - Prob. 5E Ch. 18 - Prob. 6E Ch. 18 - Prob. 7E Ch. 18 - Prob. 8E Ch. 18 - Prob. 9E Ch. 18 - Prob. 10E

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:

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:

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...

Explanation of Solution

Want to see the full answer?

Chapter 18 Solutions

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:

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...