## algorithm to find if two linked lists are intersected

June 10, 2013 Leave a comment

Two different lists will be intersected when one node of a list points to the other list’s node.

The below diagram depicts the intersection of lists

How do you solve it:

- Get the difference of the two lists (In above diagram it is 5-4=1)
- Take a pointer to the bigger list and move it as many number of times as difference between the lists calculated above (bigger list with 5 nodes is moved 1 position forward with pointer1 pointing to 2)
- Take another pointer to the second list and start comparing node of first list with the node of the second list. (Take pointer2 pointing to 18)
- Whenever nodes are same we can say lists are intersected. (they get intersected at node 4)

Code:

/** * Algorithm to detect the intersecting point of two lists * * @param s1 first list which is larger in size * @param s2 second list */ public static void getIntersectPoint(SLLIntersection s1,SLLIntersection s2){ int s1Len = s1.getLength(s1.root); int s2Len = s2.getLength(s2.root); int diff = s1Len-s2Len; System.out.println("Difference between two lists is "+diff); // Move the pointer in the larger list to diff places SLLNode s1List = s1.root; if((s1Len-s2Len)>0) { while((s1List != null) && diff>0){ s1List = s1List.next; diff--; } } // Now check the point where the lists are intersecting // by checking each node in two lists simultaneously SLLNode s2List = s2.root; while(s1List != null && s1List.next != null) { if(s1List == s2List) { System.out.println("Intersection point is at "+s1List.value); } // increment the pointers s1List = s1List.next; s2List = s2List.next; } }

Output:

List1: 1 2 3 4 5 List2: 18 19 4 5 Difference between two lists is 1 Intersection point is at 4

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