Check if two linked lists merge. If so, where?

Pavel's answer requires modification of the lists as well as iterating each list twice.

Here's a solution that only requires iterating each list twice (the first time to calculate their length; if the length is given you only need to iterate once).

The idea is to ignore the starting entries of the longer list (merge point can't be there), so that the two pointers are an equal distance from the end of the list. Then move them forwards until they merge.

lenA = count(listA) //iterates list A
lenB = count(listB) //iterates list B

ptrA = listA
ptrB = listB

//now we adjust either ptrA or ptrB so that they are equally far from the end
while(lenA > lenB):
    ptrA = ptrA->next
    lenA--
while(lenB > lenA):
    prtB = ptrB->next
    lenB--

while(ptrA != NULL):
    if (ptrA == ptrB):
        return ptrA //found merge point
    ptrA = ptrA->next
    ptrB = ptrB->next

This is asymptotically the same (linear time) as my other answer but probably has smaller constants, so is probably faster. But I think my other answer is cooler.

The following is by far the greatest of all I have seen - O(N), no counters. I got it during an interview to a candidate S.N. at VisionMap.

Make an interating pointer like this: it goes forward every time till the end, and then jumps to the beginning of the opposite list, and so on. Create two of these, pointing to two heads. Advance each of the pointers by 1 every time, until they meet. This will happen after either one or two passes.

I still use this question in the interviews - but to see how long it takes someone to understand why this solution works.