python無(wú)序鏈表刪除重復(fù)項(xiàng)的方法

#!/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):
  """
  對(duì)帶頭結(jié)點(diǎn)的無(wú)序單鏈表刪除重復(fù)的結(jié)點(diǎn)
  順序刪除：通過(guò)雙重循環(huán)直接在鏈表上進(jìn)行刪除操作
  即，外層循環(huán)用一個(gè)指針從第一個(gè)結(jié)點(diǎn)開(kāi)始遍歷整個(gè)鏈表，內(nèi)層循環(huán)從外層指針指向的下一個(gè)結(jié)點(diǎn)開(kāi)始，
  遍歷其余結(jié)點(diǎn)，將與外層循環(huán)遍歷到的的指針?biāo)傅慕Y(jié)點(diǎn)的數(shù)據(jù)域相同的結(jié)點(diǎn)刪除
  :param head: 頭指針
  :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

#!/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):
  """
  遞歸法：將問(wèn)題逐步分解為小問(wèn)題，即，對(duì)于結(jié)點(diǎn)cur，首先遞歸地刪除以cur.next為首
  的子鏈表中重復(fù)的結(jié)點(diǎn)；接著刪除以cur為首的鏈表中的重復(fù)結(jié)點(diǎn)，
  :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):
  """
  對(duì)帶頭結(jié)點(diǎn)的單鏈表刪除重復(fù)結(jié)點(diǎn)
  :param head: 鏈表頭結(jié)點(diǎn)
  :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