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This article will explain in detail how LeetCode finds the middle node of the linked list. The editor thinks it is very practical, so I share it with you as a reference. I hope you can get something after reading this article.

Given the header of a linked list, if the number of nodes in the linked list is odd, the middle node is returned, otherwise any one of the middle two nodes is returned

Class ListNode: def _ init__ (self, x): self.val = x self.next = Nonedef FindKthToTail (head): "" use the fast and slow pointer, the slow pointer takes one step, and the fast pointer takes two steps. When the fast pointer points to the tail node The node where the slow pointer is located is what you want: param head:: return: "" if not head: return None fast = slow = head while fast.next: fast = fast.next slow = slow.next if fast.next: fast = fast.next return slowdef main (): zero = ListNode (0) one = ListNode (1) two = ListNode (2) ) three = ListNode (3) four = ListNode (4) zero.next = one one.next = two two.next = three three.next = four print (FindKthToTail (zero) .val) if _ _ name__ = ='_ main__': main () this is the end of the article on "how LeetCode finds the intermediate nodes of linked lists" Hope that the above content can be helpful to you, so that you can learn more knowledge, if you think the article is good, please share it for more people to see.

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