Description:
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?
Solution:
# 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
"""
dic = {}
curr = head
i = 0
while curr != None:
if curr not in dic:
dic[curr] = I
curr = curr.next
i += 1
else:
return True
return False
Runtime: 32 ms, faster than 94.98% of Python online submissions for Linked List Cycle.
Memory Usage: 19 MB, less than 7.41% of Python online submissions for Linked List Cycle.
O(1) space solution: inspired by https://www.cnblogs.com/grandyang/p/4137187.html
Grandyang:
这道题是快慢指针的经典应用。只需要设两个指针，一个每次走一步的慢指针和一个每次走两步的快指针，如果链表里有环的话，两个指针最终肯定会相遇。
class Solution(object):
def hasCycle(self,head):
"""
:type head: ListNode
:rtype: bool
"""
slow = head
fast = head
while slow != None and fast != None and fast.next != None:
slow = slow.next
fast = fast.next.next
if slow == fast:
return True
return False
Runtime: 40 ms, faster than 66.67% of Python online submissions for Linked List Cycle.
Memory Usage: 18.2 MB, less than 33.80% of Python online submissions for Linked List Cycle.
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
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.
Note: Do not modify the linked list.
Follow-up:
Can you solve it without using extra space?
Solutions:
class Solution(object):
def detectCycle(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
dic = {}
i = 0
curr = head
while curr != None:
if curr not in dic:
dic[curr] = i
i += 1
curr = curr.next
else:
return curr
return None
Runtime: 28 ms, faster than 99.51% of Python online submissions for Linked List Cycle II.
Memory Usage: 19.1 MB, less than 5.24% of Python online submissions for Linked List Cycle II.
O(1) space solution: inspired by https://www.cnblogs.com/grandyang/p/4137302.html
class Solution(object):
def detectCycle(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
slow = head
fast = head
while slow != None and fast != None and fast.next != None:
slow = slow.next
fast = fast.next.next
if fast == slow:
break
if fast == None or fast.next == None:
return None
fast = head
while slow != fast:
slow = slow.next
fast = fast.next
return slow
Runtime: 52 ms, faster than 15.51% of Python online submissions for Linked List Cycle II.
Memory Usage: 18.2 MB, less than 56.01% of Python online submissions for Linked List Cycle II.