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160. intersection of two linked lists

admin 1月 30, 2021 0

Description

Difficulty: 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: 


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.

题意:

两个链表，要求返回交叉点。

Solution

先遍历一遍，确定二者的长度。
从较长的一个开始遍历，直到二者长度相同，然后两个链表同时向后遍历，检测每一个点是不是点是不是相等
遍历两次，时间复杂度 O(n),；
新建四个指针，空间复杂度 O(1)。

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# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class (object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        pt_a, pt_b = headA, headB
        len_a, len_b = 0, 0
        
        while pt_a:
            len_a += 1
            pt_a = pt_a.next
        while pt_b:
            len_b += 1
            pt_b = pt_b.next
        
        if len_a > len_b:
            long, short = headA, headB
        else:
            long, short = headB, headA
        
        for diff in range(abs(len_a - len_b)):
            long = long.next
        
        common = None
        while long:
            if long == short:
                common = long
                break
            else:
                long, short = long.next, short.next
        return common
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