// solution in C++ #include #include using namespace std; // simple Node class class Node { public: Node(int v) : val(v), next(NULL) {} int val; Node* next; }; // function to find intersection Node Node* findNodeWhere2ListsMeet(Node* head1, Node* head2) { // interate through first list and populate map map nodeMap; while (head1 != NULL) { nodeMap[head1] = head1->val; head1 = head1->next; } // interate through second list and try to find in map while (head2 != NULL) { if (nodeMap.find(head2) != nodeMap.end()) { return head2; } head2 = head2->next; } return NULL; } // test code (happy path only) int main() { // linked list 1 values = 1,2,3,4,5,6 Node* head1 = new Node(1); Node* n12 = new Node(2); Node* n13 = new Node(3); Node* n14 = new Node(4); Node* n15 = new Node(5); Node* n16 = new Node(6); head1->next = n12; n12->next = n13; n13->next = n14; n14->next = n15; n15->next = n16; // linked list 2 values = 11,12,13,5,6 Node* head2 = new Node(11); Node* n22 = new Node(12); Node* n23 = new Node(13); head2->next = n22; n22->next = n23; n23->next = n15; Node* intersection = findNodeWhere2ListsMeet(head1, head2); while (head1 != NULL) { cout val next; } cout val next; } cout << endl;

public static void main(String[] args) { List L1 = Arrays.asList(1,2,5,7,9,10,12,18,3); List L2 = Arrays.asList(2,3,4,8,9,9,10,11,12,18); LinkedList list1 = new LinkedList(); list1.addAll(L1); LinkedList list2 = new LinkedList(); list2.addAll(L2); System.out.println("L1 : " + list1); System.out.println("L2 : " + list2); ArrayList result = new ArrayList(); System.out.println("two linked lists meet: " + findCommon(list1, list2,result)); } public static ArrayList findCommon(LinkedList list1,LinkedList list2,ArrayList result) { if (list1.isEmpty()) return result; int pop = list1.pop(); if (list2.contains(pop)) result.add(pop); return findCommon(list1, list2, result); }