【算法日常】判断环形链表

环形链表

题目来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/linked-list-cycle-ii

目前考虑到两种解法，但都需要辅助空间， 第一种 O(n) 第二种 O(1)

第一种 借助辅助字典进行判断

将走过的节点都记录在字典中，通过查询字典的key值是否存在来确定是否有环
时间复杂度为 O(n) , 空间复杂度为 O(n)

代码如下：

# -*- 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))

输出如下：

True

第二种解法 快慢指针

# -*- 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))

输出结果

True