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  • 力扣 ——Linked List Cycle II(环形链表 II) python实现

    题目描述:

    中文:

    给定一个链表,返回链表开始入环的第一个节点。 如果链表无环,则返回 null。

    为了表示给定链表中的环,我们使用整数 pos 来表示链表尾连接到链表中的位置(索引从 0 开始)。 如果 pos 是 -1,则在该链表中没有环。

    说明:不允许修改给定的链表。

     

    英文:

    Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

    To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

    Note: Do not modify the linked list.

    # Definition for singly-linked list.
    # class ListNode(object):
    #     def __init__(self, x):
    #         self.val = x
    #         self.next = None
    
    class Solution(object):
        def detectCycle(self, head):
            """
            :type head: ListNode
            :rtype: ListNode
            """
            if head == None or head.next == None:
                return None
            slow = fast = head
            while fast and fast.next:
                slow = slow.next
                fast = fast.next.next
                if fast == slow:
                    break
            if slow == fast:
                slow = head
                while slow != fast:
                    slow = slow.next
                    fast = fast.next
                return slow
            return None

    题目来源:力扣

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  • 原文地址:https://www.cnblogs.com/spp666/p/11662012.html
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