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ERROR: type should be string, got "https://leetcode.com/problems/intersection-of-two-linked-lists/description/\r\n\r\n* algorithms\r\n* Easy (32.92%)\r\n* Source Code: 160.intersection-of-two-linked-lists.py * Total Accepted: 288.9K\r\n* Total Submissions: 873.2K\r\n* Testcase Example: '8\\n[4,1,8,4,5]\\n[5,0,1,8,4,5]\\n2\\n3'\r\n\r\nWrite a program to find the node at which the intersection of two singly linked lists begins.\r\n\r\nFor example, the following two linked lists:\r\n\r\n\r\nbegin to intersect at node c1.\r\n\r\n\r\n\r\nExample 1:\r\n\r\n\r\n```\r\nInput: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3\r\nOutput: Reference of the node with value = 8\r\nInput Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.\r\n```\r\nExample 2:\r\n\r\n\r\n```\r\nInput: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1\r\nOutput: Reference of the node with value = 2\r\nInput Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.\r\n```\r\n\r\n\r\n\r\nExample 3:\r\n\r\n\r\n```\r\nInput: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2\r\nOutput: null\r\nInput Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.\r\nExplanation: The two lists do not intersect, so return null.\r\n```\r\nNotes:\r\n- If the two linked lists have no intersection at all, return null.\r\n- The linked lists must retain their original structure after the function returns.\r\n- You may assume there are no cycles anywhere in the entire linked structure.\r\n- Your code should preferably run in O(n) time and use only O(1) memory.\r\n\r\n### C++\r\n```cpp\r\nclass Solution {\r\npublic:\r\n ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {\r\n ListNode *p, *q;\r\n\r\n p = headA;\r\n q = headB;\r\n\r\n int aCount = 0;\r\n int bCount = 0;\r\n\r\n while (p != nullptr) {\r\n ++aCount;\r\n p = p->next;\r\n }\r\n\r\n while (q != nullptr) {\r\n ++bCount;\r\n q = q->next;\r\n }\r\n\r\n p = headA;\r\n q = headB;\r\n\r\n if (aCount > bCount) {\r\n int diff = aCount - bCount;\r\n\r\n for (int i = 0;i < diff; ++i)\r\n p = p->next;\r\n }\r\n\r\n if (bCount > aCount) {\r\n int diff = bCount - aCount;\r\n\r\n for (int i = 0;i < diff; ++i)\r\n q = q->next;\r\n }\r\n\r\n while (p != nullptr && q != nullptr) {\r\n if (p == q)\r\n return p;\r\n\r\n p = p->next;\r\n q = q->next;\r\n }\r\n\r\n return nullptr;\r\n\r\n }\r\n};\r\n```\r\n\r\n### Python\r\n```python\r\nclass Solution:\r\n def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:\r\n stack1 = []\r\n stack2 = []\r\n while headA:\r\n stack1.append(headA)\r\n headA = headA.next\r\n while headB:\r\n stack2.append(headB)\r\n headB = headB.next\r\n pre = None\r\n while stack1 and stack2:\r\n if stack1[-1] is stack2.pop():\r\n pre = stack1.pop()\r\n else:\r\n return pre\r\n return pre\r\n```"