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user100 [1]
3 years ago
6

Write a method called rotate that moves the value at the front of a list of integers to the end of the list. For example, if a v

ariable called list stores the values [8, 23, 19, 7, 45, 98, 102, 4], then the call of list.rotate(); should move the value 8 from the front of the list to the back of the list, changing the list to store [23, 19, 7, 45, 98, 102, 4, 8]. If the method is called for a list of 0 elements or 1 element, it should have no effect on the list. You may neither construct any new nodes to solve this problem nor change any of the data values stored in the nodes. You must solve the problem by rearranging the links of the list.
Computers and Technology
1 answer:
polet [3.4K]3 years ago
4 0

Answer:

Explanation:

public void rotate()

{

if(front == null)

return;

ListNode current = front;

ListNode firstNode = current;

while(current.next != null)

{

current = current.next;

}

current.next = front;

front = firstNode.next;

firstNode.next = null;

}

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Two numbers are given (numbers are entered from the keyboard). If both numbers are positive, then output their sum, if both numb
AleksAgata [21]

Answer:

The program in Python is as follows:

num1 = int(input())

num2 = int(input())

if num1 >=0 and num2 >= 0:

   print(num1+num2)

elif num1 <0 and num2 < 0:

   print(num1*num2)

else:

   if num1>=0:

       print(num1**2)

   else:

       print(num2**2)

Explanation:

This gets input for both numbers

num1 = int(input())

num2 = int(input())

If both are positive, the sum is calculated and printed

<em>if num1 >=0 and num2 >= 0:</em>

<em>    print(num1+num2)</em>

If both are negative, the products is calculated and printed

<em>elif num1 <0 and num2 < 0:</em>

<em>    print(num1*num2)</em>

If only one of them is positive

else:

Calculate and print the square of num1 if positive

<em>    if num1>=0:</em>

<em>        print(num1**2)</em>

Calculate and print the square of num2 if positive

<em>    else:</em>

<em>        print(num2**2)</em>

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