2018-8-27未命名文件

Posted 2021-01-02 cerofang

面试总结

• 匹配邮箱

import re

email_list = ["[email protected]", "[email protected]", "[email protected]"]

for email in email_list:
    # ret = re.match("[w]{4,20}@163.com$", email)
    # ret = re.match(r‘^[0-9a-zA-Z_]{0,19}@[0-9a-zA-Z]{1,13}.[com,cn,net]{1,3}$‘, email)
    # ret = re.match(r‘^[w]{0,19}@[0-9a-zA-Z]{1,13}.[com,cn,net]{1,3}$‘, email)
    ret = re.match(r‘^[w]{0,19}@(?:qq|163|126).[com,cn,net]{1,3}$‘, email)
    # ret = re.match("[w]{4,20}@[163,qq]+.com$", email)
    if ret:
        print("%s 是符合规定的邮件地址,匹配后的结果是:%s" % (email, ret.group()))
    else:
        print("%s 不符合要求" % email)

• 匹配电话号码

import re

str = "15827325743"
print(re.match(r"^1[3-9]d{9}$", str).group())

• 九九乘法表

for i in range(1,10):
    for j in range(1,i+1):
        print("%s*%s=%s"%(i,j,i*j),end="   ")
    print()

print ("
".join("	".join(["%s*%s=%s" %(x,y,x*y) for y in range(1, x+1)]) for x in range(1, 10)))

• 冒泡排序

list = [1, 2, 3, 4, 5, 6, 5, 4, 3, ]


def fun(list):
    for i in range(len(list) - 1):
        for j in range(len(list) - i - 1):
            if list[j] > list[j + 1]:
                list[j], list[j + 1] = list[j + 1], list[j]


fun(list)
print(list)

• 列表生成式

print([a+b for a in ["a","b","c"] for b in["x", "y", "z"]])

• 字典推导式

daibuchong

• 斐波那契数列

i,j = 1,1
while i < 1000:
    print(i)
    i,j = j,i+j

• 两数之和

nums = [2, 7, 11, 15]

class Solution(object):
    def twoSum(self, nums, target):
        if len(nums) <= 1:
            return False
        buff_dict = {}
        for i in range(len(nums)):
            if nums[i] in buff_dict:
                return [buff_dict[nums[i]], i]
            else:
                buff_dict[target - nums[i]] = i


a1 = Solution()
print(a1.twoSum(nums,13))

• 去重列表的相关操作

#! /usr/bin/env python
# *-* coding: utf-8 *-*

a = [1, 2, 3, 4, 5, 5, 5, 3, 4]

b = list({i: a.index(i) for i in a}.keys())
print("取最大值", max(b))
print("利用字典去重后得到的结果", b)

b1_dict = {i: a.count(i) for i in a}
print("拿到的字典", b1_dict)

# 将b1这个字典按值的大小进行排序
sorted_dict = sorted(b1_dict.items(), key=lambda x: x[1], reverse=True)
print("排序的字典", sorted_dict[0:2])

# 排序（出现频率前三的）
from collections import Counter

c = Counter(a).most_common(3)
print(c)


"""
取最大值 5
利用字典去重后得到的结果 [1, 2, 3, 4, 5]
拿到的字典 {1: 1, 2: 1, 3: 2, 4: 2, 5: 3}
排序的字典 [(5, 3), (3, 2)]
[(5, 3), (3, 2), (4, 2)]
"""