[Daily] algorithm to determine circular linked list

Circular linked list

Topic Source: stay button (LeetCode)
link: https://leetcode-cn.com/problems/linked-list-cycle-ii

At present taking into account the two solutions, but they require auxiliary space, the first O (n) second O (1)

Means for determining a first auxiliary dictionary

Through the node are recorded in a dictionary, the dictionary is determined by querying whether a key value exists cycloalkyl
time complexity of O (n) , the spatial complexity is O (n)

code show as below:

# -*- coding: utf-8 -*-
# @Author   : xaohuihui
# @Time     : 19-12-6
# @File     : detect_cycled.py
# Software  : study

"""
检测环形链表
"""

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

def has_cycle(head: ListNode) -> bool:
    dict_node = dict()
    i = 0
    if head and head.next:
        while head and head.next:
            id_head = str(id(head))
            if dict_node.get(id_head) is None:
                dict_node[id_head] = i
            else:
                return True
            i += 1
            head = head.next
        return False
    else:
        return False

if __name__ == '__main__':
    # head=[3,2,0,4] pos= 1
    node1 = ListNode(3)
    node2 = ListNode(2)
    node3 = ListNode(0)
    node4 = ListNode(4)

    node1.next = node2
    node2.next = node3
    node3.next = node4
    node4.next = node2

    print(has_cycle(node1))

Output is as follows:

True

The second solution Pointer speed

# -*- coding: utf-8 -*-
# @Author   : xaohuihui
# @Time     : 19-12-6
# @File     : detect_cycled.py
# Software  : study

"""
检测环形链表
"""

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

# 第二种解法
def has_cycle(head: ListNode) -> bool:
    if head and head.next:
        fast = slow = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if fast == slow:
                return True

    else:
        return False

if __name__ == '__main__':
    # head=[3,2,0,4] pos= 1
    node1 = ListNode(3)
    node2 = ListNode(2)
    node3 = ListNode(0)
    node4 = ListNode(4)

    node1.next = node2
    node2.next = node3
    node3.next = node4
    node4.next = node2

    print(has_cycle(node1))

Output

True