题目描述：

解题思路：
如果把两个链表都遍历一遍，先遍历A，再遍历B和先遍历B再遍历A，遍历次数是一样的。
而且如果两个链表后面一样的节点，那么遍历完第一个链表后遍历第二个时，便可以开始做相等比较。
代码：

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

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        if not headA or not headB:
            return
        if headA == headB:
            return headA
        
        q = headA
        p = headB

        while q.next and p.next:
            q = q.next
            p = p.next
        
        if q.next:
            q = q.next
            p = headA
            while q.next:
                q = q.next
                p = p.next
            q = headB
            p = p.next
            
        elif p.next:
            p = p.next
            q = headB
            while p and p.next:
                q = q.next
                p = p.next
            p = headA
            q = q.next
        
        else:
            p = headA
            q = headB

        while p:
            if p == q:
                return p
            else:
                p = p.next
                q = q.next