链表常见面试题

原创已于 2022-03-10 02:08:03 修改 · 336 阅读

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#链表 #数据结构 #leetcode

于 2022-03-10 01:15:34 首次发布

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本文介绍了如何在链表中使用四种不同的方法找到中位节点：midOrUpMidNode、midOrDownMidNode、midOrUpMidPreNode和midOrDownMidPreNode，适用于链表包含至少三个节点的情况。算法详细展示了如何通过快慢指针技巧确定链表的中间位置，包括中位节点和可能的前驱节点。

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class Node:
    def __init__(self, x):
        self.value = x
        self.next = None
class LinkedListMid:
    def __init__(self):
        self.size = 0
        self.head = Node(0)

    def midOrUpMidNode(self, head):
        if head is None or head.next is None or head.next.next is None:
            return head
        # 链表有三个点或以上
        slow = head.next
        fast = head.next.next
        while fast.next is not None and fast.next.next is not None:
            slow = slow.next
            fast = fast.next.next
        return slow

    def midOrDownMidNode(self, head):
        if head is None or head.next is None:
            return head
        slow = head.next
        fast = head.next
        while fast.next is not None and fast.next.next is not None:
            slow = slow.next
            fast = fast.next.next
        return slow

    def midOrUpMidPreNode(self, head):
        if head is None or head.next is None or head.next.next is None:
            return None
        slow = head
        fast = head.next.next
        while fast.next is not None and fast.next.next is not None:
            slow = slow.next
            fast = fast.next.next
        return slow

    def midOrDownMidPreNode(self, head):
        if head is None or head.next is None:
            return None
        if head.next.next is None:
            return head
        slow = head
        fast = head.next
        while fast.next is not None and fast.next.next is not None:
            slow = slow.next
            fast = fast.next.next
        return slow

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