142-linked-list-cycle-ii
Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail’s next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.
Do not modify the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1 Output: tail connects to node index 1 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: tail connects to node index 0 Explanation: There is a cycle in the linked list, where tail connects to the first node. Example 3:
Input: head = [1], pos = -1 Output: no cycle Explanation: There is no cycle in the linked list.
My solution：
Use dict data structure to store ListNodes, when there’s a same ListNode, return it.
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# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def detectCycle(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
visited = set()
tmp = head
while tmp:
if tmp in visited:
return tmp
else:
visited.add(tmp)
tmp = tmp.next
Time complexity: O(n) Space complexity: O(n)
if not use the dict data structure, we can use fast and slow index to solve.
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# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def detectCycle(self, head):
slow = head
fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
# If there is a cycle, the slow and fast pointers will eventually meet
if slow == fast:
# Move one of the pointers back to the start of the list
slow = head
while slow != fast:
slow = slow.next
fast = fast.next
return slow
# If there is no cycle, return None
return None
Time complexity: O(n) Space complexity: O(1)