两个链表的交点

Question

There are two singly linked lists in a system. By some programming error, the end node of one of the linked list got linked to the second list, forming an inverted Y shaped list. Write a program to get the point where two linked list merge.

Here is my recursive function for that -

int intersection(Node* head_ref1 , Node* head_ref2){
    Node* temp1 = head_ref1;
    Node* temp2 = head_ref2;
    if(temp1->next == temp2->next)
        return temp1->data;
        
    return intersection(temp1->next , temp2->next);
}

But it's not giving correct answer. Please explain why I'm getting the error.

Answer 1

Node *outer_head = head_ref1;
while(outer_head -> next != null)
{

  Node *inner_head = head_ref2;
  while(inner_head -> next != null)
   {
    if(inner_head == outer_head)
    {
     //Here it is merging occured 
    }
    inner_head = inner_head -> next;
   }
  outer_head = outer_head -> next;
}

Complexity

O(n1 * n2)

Explanation

The outer loop is for each node of the first linked list and inner loop is for second linked list. In the inner loop, checking if any of nodes of the second list is same as the current node of the first linked list.

Answer 2

The code below is O(max(l1,l2)) where l1 and l2 are the length of the lists. If you want to compare iterators rather than their values change *itr1;=*itr2 to itr1!=itr2;

auto itr1 = L1.rbegin(),last1=itr1;
auto itr2 = L2.rbegin(),last2=itr2;

for (; ++itr1 != L1.rend() && ++itr2 != L2.rend(); last1++, last2++) {
        if (*itr1 != *itr2) break;
}

When it exists the loop the last common value is *last1=*last2

Edit: I see that you have a singly linked list. In that case iterate over both lists and add the elements or pointers to a stack each then keep popping the stacks until the tops are different