【Leetcode】160. Intersection of Two Linked Lists

最新推荐文章于 2023-12-11 10:18:28 发布

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#Leetcode #链表

本文介绍了三种算法，用于解决寻找两个单链表的第一个公共节点的问题。通过一次或两次遍历链表，这些方法有效地确定了是否存在交点及其位置。

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# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None
class Solution1(object):
    """
    two pass method
    in first pass get each list's length la,lb, we assume that la > lb
    the longer link list take la - lb steps first , the they set out at the same time and will meet at the same time(if have)
    """
    def getIntersectionNode(self,headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        countA,countB,pNodeA,pNodeB = 0,0,headA,headB
        while(pNodeA):
            countA += 1
            pNodeA = pNodeA.next
        while(pNodeB):
            countB += 1
            pNodeB = pNodeB.next

        if countA < countB:
            headA, headB = headB, headA
            countA, countB = countB, countA
        while(countA != countB):
            countA -=1
            headA = headA.next
        while(headA and headB and headA!=headB):
            headA = headA.next
            headB = headB.next
        return headA

class Solution2(object):
    """
    Solution1 without count but pointer
    """
    def getIntersectionNode(self,headA, headB):
        l1,l2 = headA, headB
        while(l1 and l2):
            l1,l2 = l1.next, l2.next
        if l1 is None:
            l1,l2 = l2,l1
            headA,headB = headB, headA
        while(l1):
            headA = headA.next
            l1 = l1.next
        while(headA and headB and headA != headB):
            headA = headA.next
            headB = headB.next
        return headA if headA is None else headA.val



class Solution3(object):
    """
    If there exits an intersection between headA and headB ,
    If two people start from headA and headB  and change to the other list if they arrive in then end,
    they will meet at the intersection node (because they had walk the same distance)
    if the don't have intersection , after walk through two linklist, they have experience the same distance and come be none(None == None)
    """
    def getIntersectionNode(self,headA, headB):
        l1,l2 = headA, headB
        while(l1 != l2 ):
            l1  = l1.next if l1 != None else headB
            l2  = l2.next if l2 != None else headA
        return l1