目录
- 目录
- 前言
- 题目一
- 分析
- 实现
- 题目二
- 分析
- 实现
前言
朋友到阿里面试,分享两道小题,博主比较闲就试着用 Python 解答一下,实现方式肯定是多种多样的,优劣也会各有不同,欢迎交流。
题目一
三个线程交替打印 abcabcabc…,一个打印 a,一个打印 b,一个打印 c。
分析
典型的线程同步问题,解决思路是互斥锁,三个线程通过锁来完成互斥下的协作同步。
实现
import threading
lock_a = threading.Lock()
lock_b = threading.Lock()
lock_c = threading.Lock()
def print_a(num):
if num < 0:
return
lock_a.acquire()
print('a')
lock_b.release()
print_a(num-1)
def print_b(num):
if num < 0:
return
lock_b.acquire()
print('b')
lock_c.release()
print_b(num-1)
def print_c(num):
if num < 0:
return
lock_c.acquire()
print('c')
lock_a.release()
print_c(num-1)
def main():
num = 9
thread_a = threading.Thread(target=print_a, args=(num,))
thread_b = threading.Thread(target=print_b, args=(num,))
thread_c = threading.Thread(target=print_c, args=(num,))
lock_b.acquire()
lock_c.acquire()
thread_a.start()
thread_b.start()
thread_c.start()
if __name__ == '__main__':
main()
题目二
有一个 String 类型数组 arr = { "a", "b", "d", "d", "a", "d", "a", "e", "d", "c" }
,请编码实现统计该数组中字符重复次数并由多到少的顺序对 a,b,c,d,e 重新排序输出。
分析
如果用 C 语言实现,那么姑且还是一道冒泡排序算法题。对于 Python 而言,这题就是比较单纯的熟练度考验,使用 collections 模块可以轻松实现。
实现
from collections import Counter
li1 = ["a", "b", "d", "d", "a", "d", "a", "e", "d", "c"]
print(Counter(li1))
或
from collections import defaultdict
from collections import OrderedDict
li1 = ["a", "b", "d", "d", "a", "d", "a", "e", "d", "c"]
d = defaultdict(int)
for k in li1:
d[k] += 1
print OrderedDict(sorted(d.items(), key=lambda t: t[1], reverse=True))