【leetcode】(python)160. Intersection of Two Linked Lists两个链接列表的交叉点

最新推荐文章于 2021-04-01 01:03:13 发布

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最新推荐文章于 2021-04-01 01:03:13 发布

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本文介绍了一种在两个单链表中寻找第一个相交节点的高效算法，详细解析了问题的背景、输入输出示例及解决方案，强调了O(n)的时间复杂度和O(1)的空间复杂度。

160. Intersection of Two Linked Lists Easy

Description

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

For example, the following two linked lists:
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begin to intersect at node c1.

Example

Example 1

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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

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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

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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.

题意

找到两个单链表相交的第一个相同结点

注意：最好时间复杂度为O（n）,空间复杂度为O（1）.

解题思路

直接分别对两单链表进行遍历，若相交，则肯定在某处指针相等。

code

# 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
        """
        if not headA or not headB:
            return None
        a, b = headA, headB
        while a and b and a!=b:
            a = a.next
            b = b.next
            if a == b:
                return a
            if not a:
                a = headB
            if not b:
                b = headA
        return a