1 问题
利用python实现链表的反向输出，反转单链表。
2 方法
1.程序分析:无
代码清单 1
| class Node: def __init__(self, data): self.data = data self.next = None def get_data(self): return self.data class List: def __init__(self, head): self.head = head def is_empty(self): return self.get_len() == 0 def get_len(self): length = 0 temp = self.head while temp is not None: length += 1 temp = temp.next return length def append(self, node): temp = self.head while temp.next is not None: temp = temp.next temp.next = node def delete(self, index): if index < 1 or index > self.get_len(): print("给定位置不合理") return if index == 1: self.head = self.head.next return temp = self.head cur_pos = 0 while temp is not None: cur_pos += 1 if cur_pos == index-1: temp.next = temp.next.next temp = temp.next def insert(self, pos, node): if pos < 1 or pos > self.get_len(): print("插入结点位置不合理") return temp = self.head cur_pos = 0 while temp is not Node: cur_pos += 1 if cur_pos == pos-1: node.next = temp.next temp.next =node break temp = temp.next def reverse(self, head): if head is None and head.next is None: return head pre = head cur = head.next while cur is not None: temp = cur.next cur.next = pre pre = cur cur = temp head.next = None return pre def print_list(self, head): init_data = [] while head is not None: init_data.append(head.get_data()) head = head.next return init_data if __name__=='__main__': head=Node('head') link=List(head) for i in range(10): node=Node(i) link.append(node) print(link.print_list(head)) print(link.print_list(link.reverse(head))) |
3 结语
针对此类边界条件较多的问题，提出定义类，函数的方法，通过实验，证明该方法是有效的。此方法结合熟练运用了所学的基本python知识，知识虽简单，但需要较清晰的逻辑关系来分类讨论。对验证对象条件的充分理解，是解决此题的关键。希望未来能利用更复杂的python知识解决更多问题。
