2019-08-24LeetCode160. 相交链表

10min，一次通过

class Solution(object):
    def getIntersectionNode(self, headA:ListNode, headB:ListNode):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        cnt1,cnt2=0,0
        cur=headA
        while cur:
            cnt1+=1
            cur=cur.next
        cur=headB
        while cur:
            cnt2+=1
            cur=cur.next
        while cnt1>cnt2:
            headA=headA.next
            cnt1-=1
        while cnt1<cnt2:
            headB=headB.next
            cnt2-=1
        while headA and headB!=headA: # 没有公共节点
            headB=headB.next
            headA=headA.next
        return headA

考虑到了两个都为空，没有公共节点的情况，很好，一次通过