70e74a4027
Write a predicate for checking if a linked-list contains a cycle. For additional practice, I also implemented a function that accepts a linked-list containing a cycle and returns the first element of that cycle.
70 lines
1.5 KiB
Python
70 lines
1.5 KiB
Python
def contains_cycle(node):
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"""
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Return True if the linked-list, `node`, contains a cycle.
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"""
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if not node:
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return False
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a = node
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b = node.next
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while a != b:
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a = a.next
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if b and b.next and b.next.next:
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b = b.next.next
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else:
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return False
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return True
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################################################################################
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# Bonus
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################################################################################
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def first_node_in_cycle(node):
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"""
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Given that the linked-list, `node`, contains a cycle, return the first
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element of that cycle.
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"""
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# enter the cycle
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a = node
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b = node.next
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while a != b:
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a = a.next
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b = b.next.next
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# get the length of the cycle
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beg = a
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a = a.next
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n = 1
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while a != beg:
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a = a.next
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n += 1
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# run b n-steps ahead of a
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a = node
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b = node
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for _ in range(n):
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b = b.next
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# where they intersect is the answer
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while a != b:
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a = a.next
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b = b.next
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return a
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################################################################################
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# Tests
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################################################################################
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class Node(object):
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def __init__(self, value, next=None):
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self.value = value
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self.next = next
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def __repr__(self):
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return "Node({}) -> ...".format(self.value)
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d = Node('d')
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c = Node('c', d)
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b = Node('b', c)
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a = Node('a', b)
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d.next = b
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print(first_node_in_cycle(a))
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