Intersection of Two Linked Lists

最新推荐文章于 2024-06-24 11:25:34 发布

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本文介绍了一种使用双指针的方法来找到两个链表的交集节点。通过调整两个指针的起始位置，确保它们从较短的链表开始遍历时能同时到达交集点，从而高效地解决问题。

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求两个链表的交集节点，用双指针来遍历

ps：这道题是微软的一道题，当年微软也考过，时光飞逝啊....

答案：

# 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 headA == None:
            return  headA
        if headB == None:
            return headB
            
        lengthA  = self.getLength(headA)
        lengthB = self.getLength(headB)

        a = headA
        b = headB

        if lengthA >= lengthB:
            n = lengthA - lengthB
            while n:
                a = a.next
                n -= 1
        else:
            n = lengthB - lengthA
            while n:
                b = b.next
                n -= 1
        while a != b:
                a = a.next
                b = b.next
        return a

    def getLength(self,listNode):
        p = listNode
        length = 0
        while p:
            length += 1
            p = p.next
        return length