from collections import deque #双端队列
q = deque(range(20))
q.rotate(3)#元素循环右移三位(非位运算)
print(q)
q.rotate(-3)#元素循环左移三位(非位运算)
print(q)
#Answer:deque([17, 18, 19, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16])
#Answer:deque([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19])
q.pop()#右端弹出
print(q)
#deque([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18])
q.popleft()#左端弹出
print(q)
#deque([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18])
q.remove(3) #移除第一个找到的对象
print(q)
#deque([1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18])
#q.remove(3) #移除不存在的元素报错
#print(q)
#IndexError:ValueError: deque.remove(x): x not in deque
q.extend([1,2,3,4,4]) #右端连接上新元素
print(q)
#deque([1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 1, 2, 3, 4, 4])
q.extendleft([20,193,203]) #左端连接上新元素
print(q)
#deque([203, 193, 20, 1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 1, 2, 3, 4, 4])
q.clear() #清空所有对象
print(q)
#deque([])
#gcd
def gcd(x , y):#递归log级gcd求解
if y == 0:
return x
else:
return gcd(y, x % y)
z = gcd(15,375)
print(z)
z = gcd(15981,99)
print(z)
#编写一个函数,接受一个所有元素值都不相等的整数列表 x 和一个整数 n,
#要求将值为n的元素作为支点,将列表中所有值小于n的元素全部放到n的前面,
#所有值大于n的元素放到n的后面。
#快排雏形Hhhhh
def demo(x,n):#实现方案一
t1 = [i for i in x if i < n]
t2 = [i for i in x if i > n]
return t1 + [n] + t2
x = [223,4,5356,121,63463,1212,3456,1342,456,234,45,2]
y = demo(x,1000)
print(y)
#Answer:[223, 4, 121, 456, 234, 45, 2, 1000, 5356, 63463, 1212, 3456, 1342]
def demo(x,n):#效率更高
t1 = []
t2 = []
for i in range(len(x)):
if x[i] < n:
t1.append(x[i])
else:
t2.append(x[i])
return t1 + [n] + t2
x = [223,4,5356,121,63463,1212,3456,1342,456,234,45,2]
y = demo(x,1000)
print(y)
#Answer:[223, 4, 121, 456, 234, 45, 2, 1000, 5356, 63463, 1212, 3456, 1342]
#一点点想法:如何证明快排是稳定的???