python無序鏈表刪除重復(fù)項(xiàng)的方法

更新時間：2020年01月17日 14:54:20 作者：布?xì)W不歐

這篇文章主要介紹了python無序鏈表刪除重復(fù)項(xiàng)的方法，本文給大家介紹的非常詳細(xì)，具體一定的參考借鑒價值,需要的朋友可以參考下

題目描述：

給定一個沒有排序的鏈表，去掉重復(fù)項(xiàng)，并保留原順序如： 1->3->1->5->5->7,去掉重復(fù)項(xiàng)后變?yōu)椋?->3->5->7

方法：

順序刪除
遞歸刪除

1.順序刪除

由于這種方法采用雙重循環(huán)對鏈表進(jìn)行遍歷，因此，時間復(fù)雜度為O(n**2)
在遍歷鏈表的過程中，使用了常數(shù)個額外的指針變量來保存當(dāng)前遍歷的結(jié)點(diǎn)，前驅(qū)結(jié)點(diǎn)和被刪除的結(jié)點(diǎn)，所以空間復(fù)雜度為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):
  """
  對帶頭結(jié)點(diǎn)的無序單鏈表刪除重復(fù)的結(jié)點(diǎn)
  順序刪除：通過雙重循環(huán)直接在鏈表上進(jìn)行刪除操作
  即，外層循環(huán)用一個指針從第一個結(jié)點(diǎn)開始遍歷整個鏈表，內(nèi)層循環(huán)從外層指針指向的下一個結(jié)點(diǎn)開始，
  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

結(jié)果：

在這里插入圖片描述

2.遞歸

此方法與方法一類似，從本質(zhì)上而言，由于這種方法需要對鏈表進(jìn)行雙重遍歷，所以時間復(fù)雜度為O(n**2)
由于遞歸法會增加許多額外的函數(shù)調(diào)用，所以從理論上講，該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):
  """
  遞歸法：將問題逐步分解為小問題，即，對于結(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

結(jié)果：

在這里插入圖片描述