题目1：
实现2:
def funOne(n):
ou = 0xaaaaaaaa & n
ji = 0x55555555 & n
print(bin(n), bin(ou), bin(ji))
print((ou>>1)^(ji<<1))
题目2:
实现2:
def funThree(n):
res = 0
count = 0
while n != 0:
if n >= 1:
res = res + 1 / (10**count)
n -= 1
else:
n = n * 2
count += 1
print(res)
if __name__ == "__main__":
funThree(0.3)
题目3：
这道题目应该是有点问题 这里的k应该是3的整数次倍
实现3：
def fun():
ls = [1,1,1,4,4,4,5,5,5,13,2,2,2,9,9,9]
res = []
maxLen = 0
for i in range(len(ls)):
s = transform(ls[i])
if len(s) > maxLen:
maxLen = len(s)
res.append(s)
for i in range(len(res)):
if len(res[i]) < maxLen:
for j in range(maxLen - len(res[i])):
res[i].insert(0, 0)
myRes = []
for i in range(len(res[0])):
mySum = 0
for j in range(len(res)):
mySum = (mySum + res[j][i]) % 3
myRes.append(mySum)
result = 0
myRes.reverse()
print(myRes)
for i in range(len(myRes)):
result += myRes[i] * (3 **(i))
print("最后结果为：", result)
题目4：
求n的阶乘
实现4：
def f(n):
if n == 1:
return 1
return n*f(n-1)
if __name__ == "__main__":
print(f(4))
题目5：
打印[ i , j ] 左闭右闭
实现5：
def fOne(i, j):
if i > j:
return
print(i)
fOne(i+1,j)
if __name__ == "__main__":
fOne(11, 10)
题目6：
数组求和
实现6：
def fTwo(ls):
if len(ls) == 0:
return 0
return ls[0]+fTwo(ls[1:])
if __name__ == "__main__":
print(fTwo(list(range(1, 101))))
题目7：
字符串翻转 递归的两种方式 大同小异
实现7：
def fThree(string, left, right):
if left > right:
return ""
if left == right:
return string[left]
return string[right] + fThree(string, left+1, right-1) + string[left]
def fFour(string, end):
if(end == 0):
return ""+string[end]
return string[end]+fFour(string, end-1)
if __name__ == "__main__":
print(fThree("abcde", 0, 4))
print(fFour("abcde", 4))
