题解 | #两个链表的第一个公共结点#

• 1.先分别遍历一遍a b链表，求分别的长度
• 2.让长度长的链表先走长的差值
• 3.判断相等就返回，否则返回无

#     def __init__(self, x):
#         self.val = x
#         self.next = None

#
# 
# @param pHead1 ListNode类 
# @param pHead2 ListNode类 
# @return ListNode类
#
class Solution:
    def FindFirstCommonNode(self , pHead1 , pHead2 ):
        test1 = ListNode(0)
        test2 = ListNode(0)
        test1.next,test2.next =pHead1,pHead2
        l1,l2 = test1.next,test2.next
        len1 = len2 =0
        while l1 or l2:
            if l1!=None:
                len1 = len1+1
                l1=l1.next
            if l2!= None:
                len2 =len2+1
                l2= l2.next
        l1,l2 = test1.next,test2.next
        distance = abs(len1-len2)
        while l1!=l2 :
            if l1!=None and distance==0:
                l1=l1.next
            if l2!=None and distance==0:
                l2=l2.next
            if len1>=len2 and distance!=0:
                l1 = l1.next
                distance = distance-1
            if len1<len2 and distance!=0:
                l2 = l2.next
                distance = distance-1
        if l1==l2:
            return l1
        else:
            return None
                
            
        
        # write code here