leetcode-mid-Linked list-160 Intersection of Two Linked Lists-NO

mycode 用了反转链表，所以不符合题意

参考：

思路：

1 先让长的链表先走，然后相同长度下看是否相遇

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        if not headA or not headB:
            return None
        def cal(head):
            count = 0        
            while head:
                count += 1
                head = head.next
            return count
        countA = cal(headA)
        countB = cal(headB)
        plus = countA - countB
        if plus > 0:
            while plus:
                headA = headA.next
                plus -= 1
            left = countB
        else:
            plus = abs(plus)
            while plus:
                headB = headB.next
                plus -= 1
            left = countA
        while left: #这里无论是headA还是headB都可以啦，因为两个人步伐已经一致啦
            if headA == headB:
                return headA
            headA = headA.next
            headB = headB.next
            left -= 1
        return None
                

2  让短的链表走到头后，再从长链表的头走起，这样当长链表走完后，短链表刚好在长链表上走了长度的差值的步数，所以长链表再从短链表头开始走的时候，相当于两个人起跑线相同啦

class Solution(object):
    def getIntersectionNode(self, headA, headB):
 
        if not headA or not headB:
            return None
   
        p,q = headA , headB   
 
        while p != q:   # 当p不等于q时执行下面程序
            p = headB if p is None else p1.next   # 如果p不是none，就取下一个值，是NONE就让p = headB
            q = headA if q is None else q.next    # 如果q不是none，就取下一个值，是NONE就让q = headA
            
        return p1   # p ，q相等有两种情况，一种是相交了，输出相交点，一种是不相交，输出了NONE