【leetcode❤python】 160. Intersection of Two Linked Lists

#-*- coding: UTF-8 -*-
#两种方法
#方法1：
#计算出A和B两个链表的长度分别为m、n;
#长度长的链表先走m-n步，之后再一次遍历寻找
#方法2：
#先走到一个链表的尾部，从尾部开始走；
#跳到另一个链表的头部
#如果相遇，则相遇点为重合节点
class Solution(object):
    def getLinkLenth(self,head):
        lenth=0
        while head!=None:
            lenth+=1
            head=head.next
        return lenth
        
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        # if headA==None or headB==None:return None
        dummyA=ListNode(0)
        dummyA.next=headA
        dummyB=ListNode(0)
        dummyB.next=headB
        lenA=self.getLinkLenth(headA)
        lenB=self.getLinkLenth(headB)
    
        if lenA>lenB:
            xlen=lenA-lenB
            for i in xrange(xlen):
                dummyA=dummyA.next
                
        if lenB>lenA:
           
            xlen=lenB-lenA
            
            for i in xrange(xlen):
                
                dummyB=dummyB.next
       
        while dummyB.next and dummyA.next:
            
            if dummyB.next==dummyA.next:
                return dummyA.next
            dummyB=dummyB.next
            dummyA=dummyA.next
        
        return None


# listA=[]
        listB=[]
        if headA==None or headB==None:return None
        while headA:
            listA.append(headA.val)
            headA=headA.next
        while headB:
            listB.append(headB.val)
            headB=headB.next
        
        minlen=len(listA) if len(listA)<len(listB) else len(listB)
        print listA,listB,minlen
        
        if listA[-1]!=listB[-1]:return None
        
        for i in xrange(1,minlen+1):
            print i
            if listA[-i]!=listB[-i]:
               
                return ListNode(listA[-i+1])
            if i==minlen:
               
                return ListNode(listA[-i])