题目描述：
输入两个链表，找出它们的第一个公共结点。
解题思路：
链表1长m;链表2长n
思路一：两层for循环 时间复杂度O(mn)
思路二：引入两个辅助栈
思路三：先遍历一遍得到长度差，在遍历比较
#0.首先遍历一遍两个链表，得到他们的长度m,n
#1.然后计算长度差为d,另长表先走d步，然后短表也开始遍历。
#2.当遍历到第一个相同地址时，即为第一个公共节点。
#3.时间复杂度为O(m+n)
思路三
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def FindFirstCommonNode(self, pHead1, pHead2):
# write code here
p1 = pHead1
p2 = pHead2
len1 = 0
len2 = 0
#首先遍历一遍得到链表的长度
while p1:
len1 += 1
p1 = p1.next
while p2:
len2 += 1
p2 = p2.next
diff = len1-len2
plong = pHead1
pshort = pHead2
if diff < 0:
plong = pHead2
pshort = pHead1
diff = abs(diff)
#先让长链表走几步
for i in range(diff):
plong = plong.next
#两个链表一起走
while plong and pshort:
if plong.val == pshort.val:
return plong
plong = plong.next
pshort = pshort.next
return None
