leetcode Intersection of Two Linked Lists python

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.

python code：

# 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):
　　if not headA or not headB:　　　　　　#边界条件判定
　　　　return None
　　else:
　　　　p={}
　　　　i=headA
　　　　while i:　　　　　　　　　　　　　　#遍历其中一个列表，得到一个dict（hash table）
　　　　　　p[i.val]=1
　　　　　　i=i.next
　　　　i=headB
　　　　while i:
　　　　　　if i.val in p:　　　　　　　　　　#遍历另一个列表，如果某个元素在上面得到的dict中，则返回此元素
　　　　　　　　return i
　　　　　　else:
　　　　　　　　i=i.next
　　　　　return None

other solutions given by leetcode：

There are many solutions to this problem:

• Brute-force solution (O(mn) running time, O(1) memory):

  For each node ai in list A, traverse the entire list B and check if any node in list B coincides with ai.

• Hashset solution (O(n+m) running time, O(n) or O(m) memory):
  
  Traverse list A and store the address / reference to each node in a hash set. Then check every node bi in list B: if bi appears in the hash set, then bi is the intersection node.
• Two pointer solution (O(n+m) running time, O(1) memory):
  • Maintain two pointers pA and pB initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.
  • When pA reaches the end of a list, then redirect it to the head of B (yes, B, that‘s right.); similarly when pB reaches the end of a list, redirect it the head of A.
  • If at any point pA meets pB, then pA/pB is the intersection node.
  • To see why the above trick would work, consider the following two lists: A = {1,3,5,7,9,11} and B = {2,4,9,11}, which are intersected at node ‘9‘. Since B.length (=4) < A.length (=6), pB would reach the end of the merged list first, because pB traverses exactly 2 nodes less than pA does. By redirecting pB to head A, and pA to head B, we now ask pB to travel exactly 2 more nodes than pA would. So in the second iteration, they are guaranteed to reach the intersection node at the same time.
  • If two lists have intersection, then their last nodes must be the same one. So when pA/pB reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections.

Analysis written by @stellari.