단일 체인 테이블 반전 [python 실현]

단일 체인 테이블 반전 방법

첫 번째 노드가 옮겨다니기 시작하다

끝 노드가 귀속되기 시작하다

코드는 다음과 같습니다.

class ListNode:

    def __init__(self, val):
      self.val = val
      self.next = None


class Solution:
    
    def reverse_list(self, head: ListNode) -> ListNode:
        if head == None or head.next == None:
            return head

        #  ，cur ，tmp cur.next，new_head cur 
        cur = head
        tmp = None
        new_head = None
        while cur:
            tmp = cur.next
            cur.next = new_head
            new_head = cur
            cur = tmp

        return new_head       
    
    def reverse_list2(self, head: ListNode) -> ListNode:
        if head == None or head.next == None:
            return head
        #  Node head 
        Node = None
        while head:
            p = head
            head = head.next
            p.next = Node
            Node = p
        return Node
    
    def reverse_list_recusive(self, head: ListNode) -> ListNode:
        if not head or not head.next:
            return head
        #  ， ： ， None， ， 
        NewHead = self.reverse_list_recusive(head.next)
        head.next.next = head
        head.next = None
        return NewHead
        

if __name__ == "__main__":
    n1 = ListNode(1)
    n2 = ListNode(2)
    n3 = ListNode(3)
    n4 = ListNode(4)

    n1.next = n2
    n2.next = n3
    n3.next = n4

    new_head = Solution().reverse_list(n1)
    print(new_head.val)
    print(new_head.next.val)
    print(new_head.next.next.val)
    print(new_head.next.next.next.val)

    print()

    new_head = Solution().reverse_list2(new_head)
    print(new_head.val)
    print(new_head.next.val)
    print(new_head.next.next.val)
    print(new_head.next.next.next.val)

    print()

    new_head = Solution().reverse_list_recusive(new_head)
    print(new_head.val)
    print(new_head.next.val)
    print(new_head.next.next.val)
    print(new_head.next.next.next.val)

THE END