Leetcode: Intersection of Two Linked Lists

最新推荐文章于 2025-03-18 05:59:11 发布

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最新推荐文章于 2025-03-18 05:59:11 发布

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本文介绍了一种高效算法，用于在给定的两个单链表中找到它们可能的相交节点。该算法运行时间复杂度为O(n)，且仅使用常数级别的额外内存。

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Question

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.
Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

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Solution

# 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
        """

        len1 = self.getlength(headA)
        len2 = self.getlength(headB)

        n = abs(len1-len2)
        if len1<len2:
            for dummy in range(n):
                headB = headB.next
        else:
            for dummy in range(n):
                headA = headA.next

        while headA!=headB:
            headA, headB = headA.next, headB.next

        return headA

    def getlength(self,head):
        res = 0
        while head!=None:
            res += 1
            head = head.next
        return res