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solution.cpp
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solution.cpp
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#include <bits/stdc++.h>
using namespace std;
class SinglyLinkedListNode {
public:
int data;
SinglyLinkedListNode *next;
SinglyLinkedListNode(int node_data) {
this->data = node_data;
this->next = nullptr;
}
};
class SinglyLinkedList {
public:
SinglyLinkedListNode *head;
SinglyLinkedListNode *tail;
SinglyLinkedList() {
this->head = nullptr;
this->tail = nullptr;
}
void insert_node(int node_data) {
SinglyLinkedListNode* node = new SinglyLinkedListNode(node_data);
if (!this->head) {
this->head = node;
} else {
this->tail->next = node;
}
this->tail = node;
}
};
//Method to print the data of linked list.
void print_singly_linked_list(SinglyLinkedListNode* node, string sep, ofstream& fout) {
while (node) {
fout << node->data;
node = node->next;
if (node) {
fout << sep;
}
}
}
void free_singly_linked_list(SinglyLinkedListNode* node) {
while (node) {
SinglyLinkedListNode* temp = node;
node = node->next;
free(temp);
}
}
//Explanation of Algorithm.
//1. Get count of the nodes in the first list, let count be c1.
//2. Get count of the nodes in the second list, let count be c2.
//3. Get the difference of counts d = abs(c1 – c2)
//4. Now traverse the bigger list from the first node till d nodes so that from here onwards both the lists have equal no of nodes
//5. Then we can traverse both the lists in parallel till we come across a common node. Get a common node is done by comparing the address of the nodes.
// getCount() Method counts the nodes in both the linked list.
int getCount(SinglyLinkedListNode* head){
SinglyLinkedListNode* current = head;
int count =0;
while(current != NULL){
count++;
current = current->next;
}
return count;
}
// getNode() Method returns the data of common node.
int getNode(int d,SinglyLinkedListNode* head1, SinglyLinkedListNode* head2 ){
int i;
SinglyLinkedListNode* current1 = head1;
SinglyLinkedListNode* current2 = head2;
//Traverse current1 pointer till first d nodes.
for(i=0;i<d;i++){
if(current1 == NULL) return -1;
current1 = current1->next;
}
//Now parallely traverse both the linked lists till common node is reached.
while(current1 != NULL && current2 != NULL){
if(current1 == current2) return current1->data;
current1 = current1->next;
current2 = current2->next;
}
return -1;
}
//Method to search merge point in two linked lists.
int findMergeNode(SinglyLinkedListNode* head1, SinglyLinkedListNode* head2) {
int c1 = getCount(head1);
int c2 = getCount(head2);
int d;
if(c1>c2){
d=c1-c2;
return getNode(d,head1,head2);
}
else{
d=c2-c1;
return getNode(d,head2,head1);
}
}
int main()
{
ofstream fout(getenv("OUTPUT_PATH"));
int tests;
cin >> tests;
cin.ignore(numeric_limits<streamsize>::max(), '\n');
for (int tests_itr = 0; tests_itr < tests; tests_itr++) {
int index;
cin >> index;
cin.ignore(numeric_limits<streamsize>::max(), '\n');
SinglyLinkedList* llist1 = new SinglyLinkedList();
int llist1_count;
cin >> llist1_count;
cin.ignore(numeric_limits<streamsize>::max(), '\n');
for (int i = 0; i < llist1_count; i++) {
int llist1_item;
cin >> llist1_item;
cin.ignore(numeric_limits<streamsize>::max(), '\n');
llist1->insert_node(llist1_item);
}
SinglyLinkedList* llist2 = new SinglyLinkedList();
int llist2_count;
cin >> llist2_count;
cin.ignore(numeric_limits<streamsize>::max(), '\n');
for (int i = 0; i < llist2_count; i++) {
int llist2_item;
cin >> llist2_item;
cin.ignore(numeric_limits<streamsize>::max(), '\n');
llist2->insert_node(llist2_item);
}
SinglyLinkedListNode* ptr1 = llist1->head;
SinglyLinkedListNode* ptr2 = llist2->head;
for (int i = 0; i < llist1_count; i++) {
if (i < index) {
ptr1 = ptr1->next;
}
}
for (int i = 0; i < llist2_count; i++) {
if (i != llist2_count-1) {
ptr2 = ptr2->next;
}
}
ptr2->next = ptr1;
int result = findMergeNode(llist1->head, llist2->head);
fout << result << "\n";
}
fout.close();
return 0;
}