面试题汇总——编程篇

一、排序

 1 ls=[3,2,5,1]
 2 # 方法一：
 3 def bubble(ls):
 4     for i in range(0,len(ls)):
 5         for j in range(i+1,len(ls)):
 6             if ls[i]>ls[j]:
 7                 ls[i],ls[j]=ls[j],ls[i]
 8     return ls
 9 # print(bubble(ls))
10 # 方法二：
11 print(sorted(ls,reverse=True))

 二、求最大值、最小值

1 ls=[3,2,5,1,5]
2 # 方法一
3 # print(max(ls))
4 # 方法二
5 max_val=0
6 for i in ls:
7     if i>max_val:
8         max_val=i
9 print(max_val)

 三、去重

1 ls=[3,2,3,5,1,5,5]
2 # 方法一：
3 # print(set(ls))
4 # 方法二：
5 ls2=[]
6 for i in ls:
7     if i not in ls2:
8         ls2.append(i)
9 print(ls2)

 四、统计出现的次数

1 ls=[3,2,3,5,1,5,5]
2 # 方法一：
3 d={}
4 for i in ls:
5     d[i]=ls.count(i)
6 print(d)
方法二：
ls2=list(set(ls))
for i in ls2:
ls.count(i)

 五、倒序

 1 str="tom"
 2 str2=''
 3 c = list(str)
 4 # 方法一
 5 print(sorted(list(str),reverse=False))
 6 # 方法二
 7 for i in range(len(c)):
 8   str2+=c.pop()
 9 # 方法三
10 leng=len(str)
11 leng2=len(str)-1
12 for i in range(leng):
13   str2+=c[leng2-i]

 

 六、递归

 1 p = r"D:\测试用例"
 2 
 3 import os
 4 def getFile(path):
 5     p = os.listdir(path)
 6     for i in p:
 7         f = os.path.join(path,i)
 8         if os.path.isdir(f):
 9             getFile(f)
10         else:
11             print(f)
12 getFile(p)

七、装饰器

带参数的装饰器

 1 def outer2(val):
 2     def outer(fun):
 3         def inner(*args,**kwargs):
 4             if val==1:
 5                 print("decoration1..")
 6             if val==2:
 7                 print("decoration2..")
 8             fun(*args,**kwargs)
 9         return inner
10     return outer
11 
12 @outer2(2)
13 def one():
14     print('ok...')
15 one()

 八、字典排序

 1 d = {"1":"tom","3":"Lily","2":"Jim"}
 2 # 方法一：
 3 c = sorted(d.items(),key=lambda x:x[0])
 4 # print(c)
 5 # 方法二：
 6 keys = [i for i in d.keys()]
 7 skeys = sorted(keys)
 8 vals=[]
 9 for i in skeys:
10   for k,v in d.items():
11     if i==k:
12       vals.append(v)
13 # cc = map(lambda x,y:{x:y},skeys,vals)
d={}
for i in range(len(ls)):
    d[ls[i]]=ls2[i]
print(d)
 

 九、最长的公共前缀

 1 ls=["floght","floor","floy","flower"]
 2 min_lenght = len(ls[0]) 
 3 for i in ls:
 4   if min_lenght>len(i):
 5     min_lenght=len(i)
 6 ls2=[]
 7 str2=''
 8 for i in range(min_lenght):
 9   for u in ls:
10     ls2.append(u[i])
11     if ls2.count(u[i])==min_lenght:
12       str2+=u[i]
13   ls2=[]
14 print(str2)

 

 十、递归求和

1 def get_sum(num):
2   if num<=1:
3     return 1
4   return num+get_sum(num-1)
5 print(get_sum(100))

十、递归树 

 1 menu_items = [
 2     {'id': 1, 'name': '一级菜单', 'status': 1, 'parent_id': 0},
 3     {'id': 2, 'name': '二级菜单', 'status': 1, 'parent_id': 1},
 4     {'id': 3, 'name': '三级菜单', 'status': 1, 'parent_id': 2},
 5     # 其他菜单项...
 6 ]
 7 
 8 def build_menu(menu_items, parent_id=0):
 9     menu = []
10     for item in menu_items:
11         if item['parent_id'] == parent_id:
12             child_menu = build_menu(menu_items, item['id'])
13             cc = child_menu
14             menu.append({
15                 'name': item['name'],
16                 'child': child_menu
17             })
18     return menu
19 print(build_menu(menu_items))

 

十一、找出连续的值

ls=[1,2,3,5,6,8]
ls2=[]
ls3=[]
for i in ls:
  ls2.append(i)
  if i+1 not in ls:
    if ls2[0]==ls2[-1]:
      ls3.append(ls2[0])
    else:
      ls3.append([ls2[0],ls2[-1]])
    ls2=[]
print(ls3)

 

 十二、LRU 算法

https://www.jb51.net/article/249285.htm?tdsourcetag=s_pcqq_aiomsg
https://www.jianshu.com/p/e41ec08e4aa6
 1 from collections import OrderedDict
 2 class LRUCache:
 3     def __init__(self,size):
 4         self.size = size
 5         self.dic = OrderedDict()
 6     def set(self,key,val):
 7         if key in self.dic:
 8             self.dic.pop(key)
 9         if self.size == len(self.dic):
10             self.dic.popitem(last=False)
11         self.dic.update({key:val})
12     def get(self,key):
13         val = self.dic.get(key)
14         self.dic.pop(key)
15         self.dic.update({key: val})
16         return val
17 
18 r = LRUCache(3)
19 r.set("1", "1")
20 r.set("2", "2")
21 r.set("1", "3")
22 # r.set("1", "4")
23 # r.set("1", "5")
24 print(r.dic)
25 r.get("1")
26 print(r.dic)
27 # r.set("4", "4")
28 # print(r.dic)

 

from collections import OrderedDict
class LRUCache:
    def __init__(self,size):
        self.size = size
        self.dic = OrderedDict()
    def set(self,key,val):
        if key in self.dic:
            self.dic.pop(key)
        if self.size == len(self.dic):
            self.dic.popitem(last=False)
        self.dic.update({key:val})
    def get(self,key):
        val = self.dic.get(key)
        self.dic.pop(key)
        self.dic.update({key: val})
        return val

r = LRUCache(3)
r.set("1", "1")
r.set("2", "2")
r.set("1", "3")
# r.set("1", "4")
# r.set("1", "5")
print(r.dic)
r.get("1")
print(r.dic)
# r.set("4", "4")
# print(r.dic)
posted @ 2021-12-28 09:32  东方不败--Never  阅读(33)  评论(0)  编辑  收藏  举报