python无序链表删除重复项的方法

更新时间：2020-08-21 19:06:01 作者：startmvc

题目描述：给定一个没有排序的链表，去掉重复项，并保留原顺序如：1->3->1->5->5-&

题目描述：

给定一个没有排序的链表，去掉重复项，并保留原顺序如： 1->3->1->5->5->7,去掉重复项后变为：1->3->5->7

方法：

顺序删除
递归删除

1.顺序删除

由于这种方法采用双重循环对链表进行遍历，因此，时间复杂度为O(n**2) 在遍历链表的过程中，使用了常数个额外的指针变量来保存当前遍历的结点，前驱结点和被删除的结点，所以空间复杂度为O(1)


#!/usr/bin/env python3
# -*- coding: utf-8 -*-
# @Time : 2020/1/15 20:55
# @Author : buu
# @Software: PyCharm
# @Blog ：https://blog.csdn.net/weixin_44321080
class LNode:
 def __init__(self, x):
 self.data = x
 self.next = None

def removeDup(head):
 """
 对带头结点的无序单链表删除重复的结点
 顺序删除：通过双重循环直接在链表上进行删除操作
 即，外层循环用一个指针从第一个结点开始遍历整个链表，内层循环从外层指针指向的下一个结点开始，
 遍历其余结点，将与外层循环遍历到的的指针所指的结点的数据域相同的结点删除
 :param head: 头指针
 :return:
 """
 if head is None or head.next is None:
 return
 outerCur = head.next
 innerCur = None
 innerPre = None
 while outerCur is not None:
 innerCur = outerCur.next
 innerPre = outerCur
 while innerCur is not None:
 if outerCur.data == innerCur.data:
 innerPre.next = innerCur.next
 innerCur = innerCur.next
 else:
 innerPre = innerCur
 innerCur = innerCur.next
 outerCur = outerCur.next

if __name__ == '__main__':
 i = 1
 head = LNode(6)
 tmp = None
 cur = head
 while i < 7:
 if i % 2 == 0:
 tmp = LNode(i + 1)
 elif i % 3 == 0:
 tmp = LNode(i - 2)
 else:
 tmp = LNode(i)
 cur.next = tmp
 cur = tmp
 i += 1
 print("before removeDup:")
 cur = head.next
 while cur is not None:
 print(cur.data, end=' ')
 cur = cur.next
 removeDup(head)
 print("\nafter removeDup:")
 cur = head.next
 while cur is not None:
 print(cur.data, end=' ')
 cur = cur.next

结果：

在这里插入图片描述

2.递归

此方法与方法一类似，从本质上而言，由于这种方法需要对链表进行双重遍历，所以时间复杂度为O(n**2) 由于递归法会增加许多额外的函数调用，所以从理论上讲，该方法效率比方法一低


#!/usr/bin/env python3
# -*- coding: utf-8 -*-
# @Time : 2020/1/15 21:30
# @Author : buu
# @Software: PyCharm
# @Blog ：https://blog.csdn.net/weixin_44321080
class LNode:
 def __init__(self, x):
 self.data = x
 self.next = None
def removeDupRecursion(head):
 """
 递归法：将问题逐步分解为小问题，即，对于结点cur，首先递归地删除以cur.next为首
 的子链表中重复的结点；接着删除以cur为首的链表中的重复结点，
 :param head:
 :return:
 """
 if head.next is None:
 return head
 pointer = None
 cur = head
 head.next = removeDupRecursion(head.next)
 pointer = head.next
 while pointer is not None:
 if head.data == pointer.data:
 cur.next = pointer.next
 pointer = cur.next
 else:
 pointer = pointer.next
 cur = cur.next
 return head
def removeDup(head):
 """
 对带头结点的单链表删除重复结点
 :param head: 链表头结点
 :return:
 """
 if head is None:
 return
 head.next = removeDupRecursion(head.next)
if __name__ == '__main__':
 i = 1
 head = LNode(6)
 tmp = None
 cur = head
 while i < 7:
 if i % 2 == 0:
 tmp = LNode(i + 1)
 elif i % 3 == 0:
 tmp = LNode(i - 2)
 else:
 tmp = LNode(i)
 cur.next = tmp
 cur = tmp
 i += 1
 print("before recursive removeDup:")
 cur = head.next
 while cur is not None:
 print(cur.data, end=' ')
 cur = cur.next
 removeDup(head)
 print("\nafter recurseve removeDup:")
 cur = head.next
 while cur is not None:
 print(cur.data, end=' ')
 cur = cur.next

结果：

在这里插入图片描述