use pointer. shift to left if one more leading char has been found. any unmatched char resets the pointer to first char
once you went through the entire list(first one), the pointer on the second list tells you where to concatenate that gives you O(n) where n is the length of first list On Oct 7, 3:52 am, snehal jain <[email protected]> wrote: > There are two linked list, both containing a character in each node. > > If one linked list contain characters o x e n c and second contain > characters e n c a r t a then the final linked list should contain o x > e n c a r t a i.e. if the end of one list is same as the start of > second then those characters should come only once. > > can we do it in O(n+m) where n and m are the length of list. both are > singly link list. -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
