leetcode141 Linked List Cycle

"""
Given a linked list, determine if it has a cycle in it.
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.
Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.
"""
"""
好题
解法一：快慢指针法，自己AC
"""
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution1:
    def hasCycle(self, head):
        if not head:
            return False
        slow = head
        fast = head.next
        while fast:
            if slow.val == fast.val:
                return True
            if fast.next:
                fast = fast.next
            if slow.val == fast.val:
                return True
            slow = slow.next
            fast = fast.next
        return False

"""
解法二：用set()或dict()或list()来存已经遍历过的结点,判断当前结点是否在集合中
值得注意的是,set时间最短
"""

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        if not head:
            return False
        s = set()
        cur = head
        while cur:
            if cur in s:
                return True
            s.add(cur)
            cur = cur.next
        return False
"""
dict的一些方法：
dic = {'name':'fuyong','age':29,'job':'none'}
dic.setdefault('addr','henan')
print(dic) #结果 {'addr': 'henan', 'age': 29, 'name': 'fuyong', 'job': 'none'}
dic.update({'addr':'henan'})
print(dic) #结果 {'job': 'none', 'addr': 'henan', 'age': 29, 'name': 'fuyong'}
dic.pop('job')
print(dic)  #结果为：{'age': 29, 'name': 'fuyong'}
print(dic.popitem())  #结果为('name', 'fuyong')
del dic['job']
print(dic) #结果为：{'age': 29, 'name': 'fuyong'}
dic.clear()
print(dic)   #结果为：{}
用来遍历的方法dic.keys() dic.values() dic.items()
"""