leetcode - 141-- circular linked list,

Subject description:

Given a list, the list is determined whether a ring. To show the list given ring, we integer pos connected to the end of the list to represent the position of the list (index starts from 0). If pos is -1, the ring is not on this list.

Example 1:

输入：head = [3,2,0,-4], pos = 1
输出：true
解释：链表中有一个环，其尾部连接到第二个节点。

Example 2:

输入：head = [1,2], pos = 0
输出：true
解释：链表中有一个环，其尾部连接到第一个节点。

Example 3:

输入：head = [1], pos = -1
输出：false
解释：链表中没有环。

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Problem-solving ideas:

Use double-pointer speed, fast hands each take two steps, each step by slow pointer. Traversing the above flow, if the pointer quickly come None, then there is no ring (inner annular absence of chain termination). If the fast and slow pointer as the pointer, then there is a ring

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Code:

class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution(object):
    def hasCycle(self, head):

        if head == None: return False
        slow = head
        fast = head.next
        while fast!= slow:
            if fast == None or fast.next == None or fast.next.next == None:
                return False

            slow = slow.next
            fast = fast.next.next
        return True

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Reference links:
leetcode endless chain-141-

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Topic Source:
https://leetcode-cn.com/problems/linked-list-cycle