题解 | 判断链表中是否有环 | python双指针

判断是否有环，需要两个指针，一个一次走一步，一个一次走两步，相遇说明链表有环，快指针到头结束循环代表没有换

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

#
# 
# @param head ListNode类 
# @return bool布尔型
#
class Solution:
    def hasCycle(self , head ):
        # write code here
        if not head:
            return False
        slow , fast = head , head
        while fast and fast.next:
            # 慢指针走一步，快指针走两步
            slow = slow.next
            fast = fast.next.next
            # 相遇说明有环，返回True
            if slow == fast:
                return True
        return False