141-Linked List Cycle

100-Same Tree | Links:

题意

Given head, the head of a linked list, determine if the linked list has a cycle in it. 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. Note that pos is not passed as a parameter. Return true if there is a cycle in the linked list. Otherwise, return false. Follow up: Can you solve it using O(1) (i.e. constant) memory?

Example:
Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).

思路

• 链表 + 快慢针

# 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
        """
        if not head or not head.next: return False
        slow, fast = head, head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast: return True
        return False

分析:

• Time: O(n)
• Space: O(1)