面向对象三大基本特征: 封装,继承,多态
二分法/折半 查找:若存在返回下标,若不存在,返回-1
a = [1, 2, 3, 4, 5, 7, 9, 11, 34, 43, 167, 345, 678, 899]
s = 5
t= -1
length = len(a)
start = 0
end =length -1
while end >= start:
half = (end+start)//2
if a[half] > s:
end = half
elif a[half] < s :
start = half + 1
else:
t = half
break
print(t)
九九乘法表
for i in range(1,10):
for j in range(1,i+1):
print(i,"X",j,"=",j*i,end="\t")
print()
冒泡排序
a = [3,5,7,1,4,34,678,2,167,899,43,11,9,345]
b = len(a)
for i in range(b-1,0,-1):
for j in range(i):
if a[j] > a[j+1]:
a[j],a[j+1] =a[j+1],a[j]
print(a)
1到100内的和
s = 0
for i in range(1,101):
s= s+ i
print(s)
替换
a = "我是xxx,我来自wwww"
a = a.replace("我是xxx,我来自wwww","我是小明来自上海")
print(a)
逢七过一
for i in range(1000):
if(i % 7 == 0):
print("过")
elif(i % 10 == 7):
print("过")
else:
print(i)
逢七过二
for i in range(1000):
if(i % 7 == 0):
print("过")
continue
if(i % 10 == 7):
print("过")
continue
print(i)
找出200以内的回数
#找出200以内的"回数"(即：从左向右，从右向左，都是一样的数，比如:131, 141)
def is_palindrome(n):
if n < 10:
return True
s = str(n)
for i in range(0, int(len(s) / 2)):
if s[i] == s[-i - 1]:
return True
return False
print(list(filter(is_palindrome, range(1, 201))))
# 或者
def is_palindrome(n):
s1 = str(n)
s2 = list(reversed(s1))
return list(s1) == s2
print(list(filter(is_palindrome, range(1, 201))))
