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：

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：

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