快慢指针：判断链表是否存在环的巧妙算法

2024-01-30 07:11:51

当然，我将以技术博客创作专家的身份，以独树一帜的观点来撰写一篇关于通过快慢指针判断链表是否有环的文章。

在计算机科学领域，链表是一种重要的数据结构，广泛用于各种应用中。链表中的每个节点包含一个值和一个指向下一个节点的指针。然而，在某些情况下，链表可能会形成一个环，即最后一个节点指向链表中的某个前面的节点。这种情况称为链表有环。

快速判断一个链表是否有环至关重要，因为这可以帮助我们避免陷入死循环或浪费计算资源。我们常用的方法是使用快慢指针算法。快慢指针算法的原理是：使用两个指针，一个称为快指针，另一个称为慢指针。快指针以每次跳过一个节点的速度前进，而慢指针以每次跳过一个节点的速度前进。如果链表存在环，那么快指针和慢指针最终将相遇。如果链表没有环，那么快指针将到达链表的末尾。

def has_cycle(head):
  """
  Determine if a linked list has a cycle.

  Args:
    head: The head of the linked list.

  Returns:
    True if the linked list has a cycle, False otherwise.
  """

  # Check if the linked list is empty.
  if head is None:
    return False

  # Initialize the slow and fast pointers.
  slow = head
  fast = head.next

  # Move the slow and fast pointers until they meet or the fast pointer reaches the end of the linked list.
  while slow != fast and fast is not None and fast.next is not None:
    slow = slow.next
    fast = fast.next.next

  # If the slow and fast pointers meet, then the linked list has a cycle.
  return slow == fast

代码示例

class Node:
    def __init__(self, value):
        self.value = value
        self.next = None


def has_cycle(head):
    """
    Determine if a linked list has a cycle.

    Args:
        head: The head of the linked list.

    Returns:
        True if the linked list has a cycle, False otherwise.
    """

    # Check if the linked list is empty.
    if head is None:
        return False

    # Initialize the slow and fast pointers.
    slow = head
    fast = head.next

    # Move the slow and fast pointers until they meet or the fast pointer reaches the end of the linked list.
    while slow != fast and fast is not None and fast.next is not None:
        slow = slow.next
        fast = fast.next.next

    # If the slow and fast pointers meet, then the linked list has a cycle.
    return slow == fast


# Test the has_cycle function.
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = head  # Create a cycle.

print(has_cycle(head))  # True

head = Node(1)
head.next = Node(2)
head.next.next = Node(3)

print(has_cycle(head))  # False