Sword finger offer 52. The first common node of the two linked lists

subject

Enter two linked lists , Find their first common node .

Like the two linked lists below ：

At the node c1 Start meeting .

Example 1：

 Input ：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

Enter an explanation ： The value of the intersection node is 8 （ Be careful , Cannot be if two lists intersect 0）. Starting from the respective heads , Linked list A by [4,1,8,4,5], Linked list B by [5,0,1,8,4,5]. stay A in , There is... Before the intersection node 2 Nodes ; stay B in , There is... Before the intersection node 3 Nodes .

Ideas

• The journey is the same ,A End of traversal , Traverse B;B End of traversal , Traverse A, Nodes of the same length are the same

Program

# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        A ,B = headA,headB

        while A != B :
            A = A.next if A else  headB
            B = B.next if B else headA 
        return A