【leetcode】Intersection of Two Linked Lists

题目简述：

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A: a1 → a2
↘
c1 → c2 → c3
↗
B: b1 → b2 → b3
begin to intersect at node c1.

Notes:

If the two linked lists have no intersection at all, return null.
The linked lists must retain their original structure after the function returns.
You may assume there are no cycles anywhere in the entire linked structure.
Your code should preferably run in O(n) time and use only O(1) memory.

解题思路：

首先这道题目有他的特殊性，这个特殊性就在这里的两个链表如果有交集的话，那么他们在最后面的一定都是相同的。
所以这里催生了两种想法：

1. 从后往前比较，找到最后形同的那个

2. 从前往后比较，但是这里要用到一个技巧。我们首先要去获取到这两个链表的长度，然后让长度长的那个先多走长度差的距离，最后再开始比较，第一个相同的即是。
   这里使用了第二种思路：

   Definition for singly-linked list.
   class ListNode:
   def init(self, x):
   self.val = x
   self.next = None

   class Solution:
   @param two ListNodes
   @return the intersected ListNode
   def getIntersectionNode(self, headA, headB):
   ta = ListNode(0)
   tb = ListNode(0)
   ta = headA
   tb = headB
   la = 0
   lb = 0
   while ta != None:
   la += 1
   ta = ta.next
   while tb != None:
   lb += 1
   tb = tb.next
   if la > lb :
   ta = headA
   tb = headB
   tt = la - lb
   while tt > 0:
   ta = ta.next
   tt -= 1
   while ta != None and tb != None:
   if ta == tb:
   return ta
   ta = ta.next
   tb = tb.next
   return None

        else:
            ta = headA
            tb = headB
            tt = lb -la
            while tt > 0:
                tb = tb.next
                tt -= 1
            while ta != None and tb != None:
                if ta.val == tb.val:
                    return ta
                ta = ta.next
                tb = tb.next
            return None