is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.
CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.
Most engineers make critical mistakes on their resumes -- we can fix your resume with our custom resume review service. And, we use fellow engineers as our resume reviewers, so you can be sure that we "get" what you're saying.
Our Mock Interviews will be conducted "in character" just like a real interview, and can focus on whatever topics you want. All our interviewers have worked for Microsoft, Google or Amazon, you know you'll get a true-to-life experience.
<pre lang="" line="1" title="CodeMonkey27600" class="run-this">#include<stdio.h>
- labyrinth43 June 10, 2011#include<stdlib.h>
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* Function to get the counts of node in a linked list */
int getCount(struct node* head);
/* function to get the intersection point of two linked
lists head1 and head2 where head1 has d more nodes than
head2 */
int _getIntesectionNode(int d, struct node* head1, struct node* head2);
/* function to get the intersection point of two linked
lists head1 and head2 */
int getIntesectionNode(struct node* head1, struct node* head2)
{
int c1 = getCount(head1);
int c2 = getCount(head2);
int d;
if(c1 > c2)
{
d = c1 - c2;
return _getIntesectionNode(d, head1, head2);
}
else
{
d = c2 - c1;
return _getIntesectionNode(d, head2, head1);
}
}
/* function to get the intersection point of two linked
lists head1 and head2 where head1 has d more nodes than
head2 */
int _getIntesectionNode(int d, struct node* head1, struct node* head2)
{
int i;
struct node* current1 = head1;
struct node* current2 = head2;
for(i = 0; i < d; i++)
{
if(current1 == NULL)
{ return -1; }
current1 = current1->next;
}
while(current1 != NULL && current2 != NULL)
{
if(current1 == current2)
return current1->data;
current1= current1->next;
current2= current2->next;
}
return -1;
}
/* Takes head pointer of the linked list and
returns the count of nodes in the list */
int getCount(struct node* head)
{
struct node* current = head;
int count = 0;
while (current != NULL)
{
count++;
current = current->next;
}
return count;
}
/* IGNORE THE BELOW LINES OF CODE. THESE LINES
ARE JUST TO QUICKLY TEST THE ABOVE FUNCTION */
int main()
{
/*
Create two linked lists
1st 3->6->9->15->30
2nd 10->15->30
15 is the intersection point
*/
struct node* newNode;
struct node* head1 =
(struct node*) malloc(sizeof(struct node));
head1->data = 10;
struct node* head2 =
(struct node*) malloc(sizeof(struct node));
head2->data = 3;
newNode = (struct node*) malloc (sizeof(struct node));
newNode->data = 6;
head2->next = newNode;
newNode = (struct node*) malloc (sizeof(struct node));
newNode->data = 9;
head2->next->next = newNode;
newNode = (struct node*) malloc (sizeof(struct node));
newNode->data = 15;
head1->next = newNode;
head2->next->next->next = newNode;
newNode = (struct node*) malloc (sizeof(struct node));
newNode->data = 30;
head1->next->next= newNode;
head1->next->next->next = NULL;
printf("\n The node of intersection is %d \n",
getIntesectionNode(head1, head2));
getchar();
}</pre><pre title="CodeMonkey27600" input="yes">
</pre>