剑指offer python版 两个链表的第一个公共结点

class ListNode(object):
    def __init__(self,x):
        self.val=x
        self.next=None
        
class Link(object):
    def __init__(self,values=None):
        self.nodes=self.set_link(values) if  values else None
        
    def get_link(self):
        return self.nodes
    def set_link(self,values):
        if not values :
            return False
        head=ListNode(0)
        move=head
        
        
        try:
            for i in values:
                tmp=ListNode(i)
                move.next=tmp
                move=move.next
                
        except Exception as e:
            print(e)
            
        return head.next
def aa(link1, link2):
    if not link1 or not link2:
        return None
    length1 = length2 = 0
    move1, move2 = link1, link2
    while move1:  # 获取链表长度
        length1 += 1
        move1 = move1.next
    while move2:
        length2 += 1
        move2 = move2.next
    while length1 > length2:  # 长链表先走多的长度
        length1 -= 1
        link1 = link1.next
    while length2 > length1:
        length2 -= 1
        link2 = link2.next
    while link1:  # 链表一起走
        if link1 == link2:
            return link1
        link1, link2 = link1.next, link2.next
    return None        
        






a=Link([1,2,3,3,4,4])
b=Link([3,6,3,2])

m=a.get_link()
q=b.get_link()

print(aa(m,q))
        
        
def aa(head1,head2):
    if not head1 or not head2:
        return False
    link1=0
    link2=0
    move1=head1
    move2=head2
    while move1:
        link1 +=1
        move1=move1.next
    while move2:
        link2 +=1
        move2=move2.next
        
    m=link1-link2
    if m>0:
        while m>0:
            head1=head1.next
            m -=1
    else:
        while m<0:
            head2=head2.next
            m +=1
            
    while head1:
        if head1==head2:
            return head1
        head1=head1.next
        head2=head2.next
        
    return None