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

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