JonnyF--Intersection of Two Linked Lists

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：

       可以将A,B两个链表看做两部分，交叉前与交叉后。交叉后的长度是一样的，因此交叉前的长度差即为总长度差。只要去除这些长度差，距离交叉点就等距了。为了节省计算，在计算链表长度的时候，顺便比较一下两个链表的尾节点是否一样，若不一样，则不可能相交，直接可以返回NULL。

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