6 - Easy - 环形链表

给定一个链表，判断链表中是否有环。

进阶：
你能否不使用额外空间解决此题？

快慢指针法，如果有环，总会相遇

# 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
        """
        slow, fast = head, head
        while slow and fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                return True
        return False

每经过一个节点，都卷回头指针，因此如果有环，必会碰到头指针
循环次数比上一个少

# 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
        """
        curr = head
        while curr is not None:
            nextnode = curr.next
            if nextnode is head:
                return True
            curr.next = head
            curr = nextnode
        return False