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We can use TwoPointerTwo PointerTwoPointer algorithm to find intersected node.

Consider the following Linked List(LL) Structure:

A-->B-->C-->D--|

...............|-->E-->F-->G

.......H-->L-->|

Here, head of first LL is A and head of second LL is H Both intersect at node E

Let distance from node A to D is x and distance from node H to L is y. Distance from node E to G is common and it is z.

During first traversal of LL1 and LL2, they will cover distance: for LL1 pointer --> x+z for LL2 pointer --> y+z

When the pointers inerchange their position in second traversal then distance covered till D and L are same : for LL1 pointer --> x+z+y for LL2 pointer --> y+z+x

if they will intersect then there exist a node that is common.

• Time Complexity: O(n+m)O(n+m)O(n+m)
• Space Complexity: O(1)O(1)O(1)

 public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    ListNode node1 = headA, node2 = headB;
    int count = 0; // maintain count of traversal
    while(count<2){
        if(node1==node2) return node1;
        if(node1.next==null) ++count; // if no match after traversing twice then LL will not intersect.
        node1 = node1.next==null?headB:node1.next;
        node2 = node2.next==null?headA:node2.next;
    }

    return null;
}