文章目录
- 第一部分 Python之禅
- 第二部分 时间复杂度分析
- 2.1 常见时间复杂度
- 2.2 三集不相交问题
- 2.3 元素唯一性问题
- 2.4 第n个斐波那契数
- 2.5 最大盛水容器
- 2.6 是不是时间复杂度低就一定好?
- 2.7 影响运算速度的因素
第一部分 Python之禅
import this
The Zen of Python, by Tim Peters
Beautiful is better than ugly.
Explicit is better than implicit.
Simple is better than complex.
Complex is better than complicated.
Flat is better than nested.
Sparse is better than dense.
Readability counts.
Special cases aren't special enough to break the rules.
Although practicality beats purity.
Errors should never pass silently.
Unless explicitly silenced.
In the face of ambiguity, refuse the temptation to guess.
There should be one-- and preferably only one --obvious way to do it.
Although that way may not be obvious at first unless you're Dutch.
Now is better than never.
Although never is often better than *right* now.
If the implementation is hard to explain, it's a bad idea.
If the implementation is easy to explain, it may be a good idea.
Namespaces are one honking great idea -- let's do more of those!
- Beautiful is better than ugly
整齐、易读胜过混乱、晦涩
- Simple is better than complex
简约胜过复杂
- Complex is better than complicated
复杂胜过晦涩
- Flat is better than nested
扁平胜过嵌套
- Now is better than never.
Although never is often better than right now.
理解一:先行动起来,编写行之有效的代码,不要企图一开始就编写完美无缺的代码
理解二:做比不做要好,但是盲目的不加思考的去做还不如不做
- If the implementation is hard to explain, it’s a bad idea.
If the implementation is easy to explain, it may be a good idea.
如果方案很难解释,很可能不是有一个好的方案,反之亦然
- 个人感悟
- 首先要行动起来,编写行之有效的代码;
- 如果都能解决问题,选择更加简单的方案;
- 整齐、易读、可维护性、可扩展性好;
- 强壮、鲁棒性好,抗扰动,容错,泛化能力好;
- 响应速度快,占用空间少。
- 有些时候,鱼和熊掌不可兼得,根据实际情况进行相应的取舍
第二部分 时间复杂度分析
2.1 常见时间复杂度
def one(x):
"""常数函数"""
return np.ones(len(x))
def log(x):
"""对数函数"""
return np.log(x)
def equal(x):
"""线性函数"""
return x
def n_logn(x):
"""nlogn函数"""
return x*np.log(x)
def square(x):
"""平方函数"""
return x**2
def exponent(x):
"""指数函数"""
return 2**x
import matplotlib.pyplot as plt
plt.style.use("seaborn-whitegrid")
t = np.linspace(1, 20, 100)
methods = [one, log, equal, n_logn, square, exponent]
method_labels = ["$y = 1$", "$y = log(x)$", "$y = x$", "$y = xlog(x)$", "$y = x^2$", "$y = 2^x$"]
plt.figure(figsize=(12, 6))
for method, method_label in zip(methods, method_labels):
plt.plot(t, method(t), label=method_label, lw=3)
plt.xlim(1, 20)
plt.ylim(0, 40)
plt.legend()
plt.show()
- 我们的最爱:常数函数和对数函数
- 勉强接受:线性函数和nlogn函数
- 难以承受:平方函数和指数函数
2.2 三集不相交问题
- 问题描述: 假设有A、B、C三个序列,任一序列内部没有重复元素,欲知晓三个序列交集是否为空
import random
def creat_sequence(n):
A = random.sample(range(1, 1000), k=n)
B = random.sample(range(1000, 2000), k=n)
C = random.sample(range(2000, 3000), k=n)
return A, B, C
A, B, C = creat_sequence(100)
def no_intersection_1(A, B, C):
for a in A:
for b in B:
for c in C:
if a == b == c:
return False
return True
%timeit no_intersection_1(A, B, C)
36.7 ms ± 2.12 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
def no_intersection_2(A, B, C):
for a in A:
for b in B:
if a == b:
for c in C:
if a == c:
return False
return True
%timeit no_intersection_2(A, B, C)
301 µs ± 37.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
import time
res_n_3 = []
res_n_2 = []
for n in [10, 20, 100]:
A, B, C = creat_sequence(n)
start_1 = time.time()
for i in range(100):
no_intersection_1(A, B, C)
end_1 = time.time()
for i in range(100):
no_intersection_2(A, B, C)
end_2 = time.time()
res_n_3.append(str(round((end_1 - start_1)*1000))+"ms")
res_n_2.append(str(round((end_2 - end_1)*1000))+"ms")
print("{0:<23}{1:<15}{2:<15}{3:<15}".format("方法", "n=10", "n=20", "n=100"))
print("{0:<25}{1:<15}{2:<15}{3:<15}".format("no_inte rsection_1", *res_n_3))
print("{0:<25}{1:<15}{2:<15}{3:<15}".format("no_intersection_2", *res_n_2))
方法 n=10 n=20 n=100
no_inte rsection_1 6ms 42ms 4001ms
no_intersection_2 0ms 1ms 24ms
2.3 元素唯一性问题
def unique_1(A):
for i in range(len(A)):
for j in range(i+1, len(A)):
if A[i] == A[j]:
return False
return True
def unique_2(A):
A_sort = sorted(A)
for i in range(len(A_sort)-1):
if A[i] == A[i+1]:
return False
return True
import time
import random
res_n_2 = []
res_n_log_n = []
for n in [100, 1000]:
A = list(range(n))
random.shuffle(A)
start_1 = time.time()
for i in range(100):
unique_1(A)
end_1 = time.time()
for i in range(100):
unique_2(A)
end_2 = time.time()
res_n_2.append(str(round((end_1 - start_1)*1000))+"ms")
res_n_log_n.append(str(round((end_2 - end_1)*1000))+"ms")
print("{0:<13}{1:<15}{2:<15}".format("方法", "n=100", "n=1000"))
print("{0:<15}{1:<15}{2:<15}".format("unique_1", *res_n_2))
print("{0:<15}{1:<15}{2:<15}".format("unique_2", *res_n_log_n))
方法 n=100 n=1000
unique_1 49ms 4044ms
unique_2 1ms 21ms
2.4 第n个斐波那契数
- 问题描述:a(n+2) = a(n+1) + a(n)
def bad_fibonacci(n):
if n <= 1:
return n
else:
return bad_fibonacci(n-2)+ bad_fibonacci(n-1)
def good_fibonacci(n):
i, a, b = 0, 0, 1
while i < n:
a, b = b, a+b
i += 1
return a
%timeit bad_fibonacci(10)
20.6 µs ± 1.15 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit good_fibonacci(10)
875 ns ± 24.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
2.5 最大盛水容器
- 问题描述:给定n个非负整数a1,a2,…,an,每个数代表坐标中的一个点(i, ai)。在坐标内画n条垂直线,垂直线i的两个端点分别为(i, ai)和(i, 0)。找出其中的两条线,使得它们与x轴共同构成的容器可以容纳最多的水。
def max_area_double_cycle(height):
"""暴力穷举双循环"""
i_left, i_right, max_area = 0,0,0
for i in range(len(height)-1):
for j in range(i+1, len(height)):
area = (j-i) * min(height[j], height[i])
if area > max_area:
i_left, i_right, max_area = i, j, area
return i_left, i_right, max_area
height = np.random.randint(1, 50, size=10)
print(height)
print(max_area_double_cycle(height))
[10 11 41 26 2 44 26 43 36 30]
(2, 8, 216)
import matplotlib.pyplot as plt
plt.bar(range(10), height, width=0.5)
plt.xticks(range(0, 10, 1))
plt.show()
- 双向指针
一头一尾两个指针,它们所构成的面积,是它们中矮的那个的高度,乘以它们的距离。
这个面积,是它们中矮的那个与其它所有柱子所能构成的最大面积。
求出面积并记录,然后矮的那个指针向中间走一步,相当于把这个柱子舍弃。
直到指针相遇,共走了n-2步。
def max_area_bothway_points(height):
"""双向指针法"""
i = 0
j = len(height)-1
i_left, j_right, max_area=0, 0, 0
while i < j:
area = (j-i) * min(height[i], height[j])
if area > max_area:
i_left, j_right, max_area = i, j, area
if height[i] == min(height[i], height[j]):
i += 1
else:
j -= 1
return i_left, j_right, max_area
double_cycle = []
bothway_points = []
for n in [5, 50, 500]:
height = np.random.randint(1, 50, size=n)
start_1 = time.time()
for i in range(100):
max_area_double_cycle(height)
end_1 = time.time()
for i in range(100):
max_area_bothway_points(height)
end_2 = time.time()
double_cycle.append(str(round((end_1 - start_1)*1000))+"ms")
bothway_points.append(str(round((end_2 - end_1)*1000))+"ms")
print("{0:<15}{1:<15}{2:<15}{3:<15}".format("方法", "n=5", "n=50", "n=500"))
print("{0:<13}{1:<15}{2:<15}{3:<15}".format("暴力循环", *double_cycle))
print("{0:<13}{1:<15}{2:<15}{3:<15}".format("双向指针", *bothway_points))
方法 n=5 n=50 n=500
暴力循环 3ms 97ms 7842ms
双向指针 2ms 8ms 56ms
2.6 是不是时间复杂度低就一定好?
- 举个极端的例子:100000n VS 0.00001n2,
n特别大的时候确实是0.00001n2更大,但是有时需要处理的数据量比较小,0.00001n2反而更小
2.7 影响运算速度的因素