力扣-41-环形链表

问题:

# 给定一个链表,判断链表中是否有环。 
#
# 如果链表中有某个节点,可以通过连续跟踪 next 指针再次到达,则链表中存在环。 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。注意:pos 不作为参数进行传递,仅仅是为了标识链表的实际情况。
#
# 如果链表中存在环,则返回 true 。 否则,返回 false 。

方法一:暴力,hash/set

# leetcode submit region begin(Prohibit modification and deletion)
# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        res = set()
        while head and head.next:
            if head not in res:
                res.add(head)
            else:
                return True
            head = head.next
        return False
# leetcode submit region end(Prohibit modification and deletion)

方法二:双指针法

# leetcode submit region begin(Prohibit modification and deletion)
# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        if not head or not head.next:
            return False
        slow = head
        fast = head.next
        while slow != fast:
            if not fast or not fast.next:
                return False
            slow = slow.next
            fast = fast.next.next
        return True
# leetcode submit region end(Prohibit modification and deletion)
# leetcode submit region begin(Prohibit modification and deletion)
# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        if not head or not head.next:
            return False
        slow = head
        fast = head.next
        while slow != fast:
            if not fast or not fast.next:
                return False
            slow = slow.next
            fast = fast.next.next
        return True
# leetcode submit region end(Prohibit modification and deletion)

 

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