Answer / coder_1

Create two pointers, each set to the start of the list.
Update each as follows:

while (pointer1) {
pointer1 = pointer1->next;
pointer2 = pointer2->next; if (pointer2)
pointer2=pointer2->next;
if (pointer1 == pointer2) {
print (\"circular\n\");
}
}

If a list is circular, at some point pointer2 will wrap
around and be either at the item just before pointer1, or
the item before that. Either way, it?s either 1 or 2 jumps
until they meet.

Answer / pratyu

node *ptr1,*ptr2;
ptr1=head;
ptr2=head->next;
while(ptr2)
{
if(ptr1==ptr2)
printf("circular linked list");

ptr2=ptr2->next;
}
printf("not a circular linked list");

Answer / priya

This is the function to check if the linklist is circular:

bool CLinklist::ifcircular()
{
node *ptr1,*ptr2;
ptr1 = front;
ptr2 = front;

while(ptr1)
{
ptr1 = ptr1->link;
ptr2 = ptr2->link;

if(ptr2)
ptr2 = ptr2->link;

if(ptr1 == ptr2)
return 1;
}
}

Answer / shruti

consider home pointer as the starting pointer of the linked
list.
consider temp as the temporary pointer.


temp = home;
while(temp != NULL)
{
if(temp -> next == start)
{
flag = 1;
break;
}
else
flag = 0;
temp = temp -> next;
}

if(flag == 1)
printf("Circular");
else
printf("Not circular");

Answer / nirwal

hi All

specialy the two pointer solution techies


what about the loop in the link list ?

how you differentiate b/w circular list and list having the
loop?

Think again...........

Answer / abhi

while(q->next->next=NULL)
{
p=p->next;
q=q->next->next;
if(p==q)
{
printf("loof is find");
break;
}
}

Answer / shahid khan abbasi

bool hasCircle(List l)
{
Iterator i = l.begin(), j = l.begin();
while (true) {
// increment the iterators, if either is at the end,
you're done, no circle
if (i.hasNext()) i = i.next(); else return false;

// second iterator is travelling twice as fast as first
if (j.hasNext()) j = j.next(); else return false;
if (j.hasNext()) j = j.next(); else return false;

// this should be whatever test shows that the two
// iterators are pointing at the same place
if (i.getObject() == j.getObject()) {
return true;
}
}
}

Answer / burmeselady

start from pointer1
check all node's next pointer is null or not starting from
pointer1 until pointer1
if a node next pointer is null=> it is not circular.
else it is circular

Answer / ishan

by checking whether link field in the last node contains
the address of the first node.

Answer / lohitha

node *ptr1,*ptr2;
ptr1=head;
ptr2=head->next;
while(ptr2)
{
if(ptr1==ptr2)
{
printf("circular linked list");
break;
}
ptr2=ptr2->next;
}
printf("not a circular linked list");

Write a routine that prints out a 2-D array in spiral order

3 Answers   Microsoft,

void main() { int i; char a[]="\0"; if(printf("%s\n",a)) printf("Ok here \n"); else printf("Forget it\n"); }

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main() { char str1[] = {‘s’,’o’,’m’,’e’}; char str2[] = {‘s’,’o’,’m’,’e’,’\0’}; while (strcmp(str1,str2)) printf(“Strings are not equal\n”); }

1 Answers  

Hi, i have a project that the teacher want a pyramid of numbers in C# or java...when we click a button...the pyramid should be generated in a listbox/or JtextArea...and the pyramid should have the folowing form: 1 232 34543 4567654 567898765 67890109876 7890123210987 890123454321098 90123456765432109 0123456789876543210 Plz help with codes...didn't find anything on the net.

0 Answers  

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String reverse with time complexity of n/2 with out using temporary variable.

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2 Answers  

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How can you relate the function with the structure? Explain with an appropriate example.

0 Answers  

What are segment and offset addresses?

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