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linked-list-cycle.py
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74 lines (47 loc) · 1.61 KB
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'''
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.
Follow up:
Can you solve it using O(1) (i.e. constant) memory?
'''
# 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
"""
# 判断单链表中有没有环,等价于判断它有没有相同节点。
# Approach one 快慢指针检测相同点 O(n) , O(1)
# fast = slow = head
# while fast and fast.next:
# fast = fast.next.next
# slow = slow.next
# if fast and slow.val == fast.val:
# return True
# return False
# Approach two 用字典
dic = {}
while head:
if head in dic:
return True
else:
dic[head] = 1
head = head.next
return False