求两个链表的交点

# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

# 将链表反转, 然后进行逆序查找
class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        if headA is None or headB is None:
            return None

        head1 = self.reverseList(headA)
        head2 = self.reverseList(headB)
        # iterator pointer
        it1 = head1
        it2 = head2
        if it1.val != it2.val:
            return None
        it0 = it1
        while it1 and it2:
            if it1.val == it2.val:
                it0 = it1
                it1 = it1.next
                it2 = it2.next
            else:
                return it0
        return it0

    def reverseList(self, head):
        # 边界处理
        if head is None:
            return None
        if head.next is None:
            return head

        #  从第二个节点开始处理
        prev = head
        cur = head.next

        while cur is not None:
            future = cur.next
            cur.next = prev
            prev = cur
            cur = future
        # 第一个节点
        head.next = None
        return prev

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