Ac solution code

  • 15


    1. When pointer X in shorter list reaches the end, pointer Y in the longer list will have len(longer) - len(shorter) left. Put pointer X to the head of the longer list, then when Y reaches its end, X already traveled len(longer)-len(shorter). Then put Y to the head of shorter list.

    2. Now X and Y have the same distance to the end:
      1). If has intersection, intersection is the first node where X = Y
      2). If no intersection, termination case is X = Y = null, where they reach end together (as X, Y have the same distance to end)

    Runtime complexity = O(m + n)

    m = len(longer), n = len(shorter):

    step 1: uses m time
    step 2: uses n time


    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    	if (headA == null || headB == null) 
    		return null;
    	ListNode curA = headA, curB = headB;
    	while (curA != curB) {
    		curA = (curA == null) ? headB :;
    		curB = (curB == null) ? headA :;			
    	return curA;

  • 0

    nice solution!

  • 1

    Won't this go in infinite loop if there is no cycle?

  • 0

    super nice solution!

  • 0

    Yes, it seems it will contains the loop.

  • 0

    no, if there is no cycle, then the 2 pointers will reach the nullptr at the same time.

  • 0

    No. You can view two linkedlist connect at the null node in the end of two list. Two pointer run through same distance, so they will definite merge at that point.

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