160. Intersection of Two Linked Lists（easy） 

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

For example, the following two linked lists:

begin to intersect at node c1.

 

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output:
 
 Reference of the node with value = 8
Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

 

Example 2:

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

 

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

 

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.

1 O(n) memory ,O(n) time

使用list儲存並遍歷，非常慢

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

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        n1 ,n2 = 0,0
        h1,h2 = headA,headB
        list1 = [];list2 = []
        while(h1):
            n1 += 1
            list1.append(h1)
            h1 = h1.next
        while(h2):
            n2 += 1
            list2.append(h2)
            h2 = h2.next
            
        for node in list1:
            if node in list2:return node

2  O(1)memory O(n)time,從後面長度相同的地方開始遍歷判斷

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

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        h1,h2 = headA,headB
        n1 ,n2 = 0,0
        while(h1):
            n1 += 1
            h1 = h1.next
        while(h2):
            n2 += 1
            h2 = h2.next
        n = min(n1,n2)
        h1,h2 = headA,headB
        if n1 > n2:
            for i in range((n1 - n2)):
                h1 = h1.next
        elif n2 > n1:
            for i in range((n2 - n1)):
                h2 = h2.next
        while(h1 != None and h2 != None):
            if h1 == h2 :return h1
            h1 = h1.next
            h2 = h2.next