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142. Linked List Cycle II.py
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142. Linked List Cycle II.py
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"""
142. Linked List Cycle II
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.
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.
Follow up:
Can you solve it without using extra space?
"""
# Definition for singly-linked list.
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
"""
使用快慢指针
"""
class Solution(object):
def meet(self, slow, fast):
while fast and fast.next:
fast = fast.next.next
slow = slow.next
if fast == slow:
return fast
return None
def detectCycle(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
slow = head
fast = head
fast = self.meet(slow, fast)
if fast:
while fast != slow:
fast = fast.next
slow = slow.next
return slow
else:
return None
if __name__ == '__main__':
node1 = ListNode(1)
node2 = ListNode(2)
node3 = ListNode(3)
node4 = ListNode(4)
node1.next = node2
node2.next = node3
node3.next = node4
node4.next = node1
solution = Solution()
print(solution.detectCycle(node1))