# Sword refers to Offer 52. The first common node of two linked lists

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### 1. Title 📑

Enter two linked lists to find their first common node.

The intersection begins at node C1.

Example 1:

Enter: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3

Output: Reference of the node with value = 8

Input explanation: The value of the intersection node is 8 (note that it cannot be 0 if two lists intersect). Starting from their respective table headers, linked list A is [4,1,8,4,5] and linked list B is [5,0,1,8,4,5]. In A, there are two nodes before the intersecting node; In B, there are three nodes before the intersection node.

Example 2:

Enter: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1

Output: Reference of the node with value = 2

Input explanation: The value of the intersection node is 2 (note that it cannot be 0 if two lists intersect). Starting from their respective table headers, linked list A is [0,9,1,2,4] and linked list B is [3,2,4]. In A, there are three nodes before the intersecting node; In B, there is 1 node before the intersecting node.

Example 3:

Enter: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2

Output: null

Input explanation: from the respective table headers, linked list A is [2,6,4] and linked list B is [1,5]. Because the two linked lists do not intersect, intersectVal must be 0, and skipA and skipB can be arbitrary values.

Explanation: The two lists do not intersect, so null is returned.

Note:

• If two lists do not intersect, null is returned.
• After the result is returned, both lists must remain in their original structure.
• It can be assumed that there are no loops in the entire list structure.
• The program meets the O(n) time complexity as far as possible and uses only O(1) memory.
• Question 160 is the same as that in the main station: leetcode-cn.com/problems/in…

### 2. Ideas 🧠

Method 1: double pointer

In general, it is very convenient to insert and delete linked lists with double Pointers, but it is very tedious to insert and delete arrays in a certain location.

1. The length of linked list A is L1+C, and that of linked list B is L2+C, where C is the length of the common part of the two lists.
2. After list A has taken L1+C steps, go back to the start of list B and take L2 steps. After list B has gone L2+C, go back to the start of list A and go L1. The list meets when the number of steps is L1+L2+C.
3. The list meets when the number of steps is L1+L2+C.
• If the two listsThere areCommon tail (i.e. C > 0) : pointer`A` , `B`It also points to a public node`node`
• If the two listsThere is noCommon tail (i.e. C = 0) : pointer`A` , `B`Also point to null.

Less nonsense ~~~~~ on the code!

### 3. Code: 👨💻

Commit AC for the first time

``````/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; *} *} */
public class Solution {
while(A ! = B) { A = A ==null ? headB : A.next;
B = B == null ? headA : B.next;
}
returnA; }}Copy the code``````

Time complexity: O(m + n) m represents the length of l1 and n represents the length of L2

Space complexity: O(1)

### 4, summarize

The biggest difficulty of this topic is whether to understand the structure of the linked list, understand the double pointer, for beginners double pointer is a very headache.

### 5. Related operations of linked lists

18. Remove list nodes – nuggets

Leetcode 21. Merge two ordered lists – nuggets

Leetcode 83. Remove duplicate elements from sorted lists – nuggets

22. The k node from the bottom of the list – nuggets

Sword finger Offer 24. Reverse the list – nuggets

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52. The first common node of two linked lists