I tried LeetCode every day 160. Intersection of Two Linked Lists (Python, Go)

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leetcode.com This is the practice of coding interviews for software developers. A total of more than 1,500 coding questions have been posted, and it seems that the same questions are often asked in actual interviews.

Introduction to golang + algorithm I will solve it with go and Python to strengthen the brain.

36th question (question 160)

160. Intersection of Two Linked Lists

the issue's details

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

begin to intersect at node c1.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input 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 [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Example 2:

Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input 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 [1,9,1,2,4]. 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.

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input 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.
Explanation: The two lists do not intersect, so return null.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Each value on each linked list is in the range [1, 10^9].
• Your code should preferably run in O(n) time and use only O(1) memory.

Way of thinking

1. Make a copy of each to proceed with headA and headB (p, q).
2. Substitute headB when p goes to the end, and assign headA when q goes to the end.
3. Loop until p and q are equal or both reach the end.
4. Returns null if it's terminating, and that node if it's equal.

Answer code

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        if not headA or not headB:
            return None
        
        p,q = headA , headB 
        while p and q and p != q :
            p = p.next
            q = q.next
            if p == q :
                return q
            if not p :
                p = headB
            if not q :
                q = headA
        return p

--I'll write it in Go too!

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func getIntersectionNode(headA, headB *ListNode) *ListNode {
	if headA == nil || headB == nil {
		return nil
	}
	A := headA
	B := headB

	for A != B {
		if A == nil {
			A = headB
		} else {
			A = A.Next
		}
		if B == nil {
			B = headA
		} else {
			B = B.Next
		}
	}
	return A
}