【leetcode】Intersection of Two Linked Lists

題目簡述：

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.

解題思路：

首先這道題目有他的特殊性，這個特殊性就在這裏的兩個鏈表若是有交集的話，那麼他們在最後面的必定都是相同的。
因此這裏催生了兩種想法：

1. 從後往前比較，找到最後形同的那個

2. 從前日後比較，可是這裏要用到一個技巧。咱們首先要去獲取到這兩個鏈表的長度，而後讓長度長的那個先多走長度差的距離，最後再開始比較，第一個相同的便是。
   這裏使用了第二種思路：

   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