1.分別使用遞迴、迴圈和生成器求菲波那切數列
遞迴：def fib_rec(n):
if n == 0:
return 1
if n == 1:
return 1
return fib_rec(n-1) + fib_rec(n-2)
fib_rec(5)
迴圈：def fib_loop(n):
lst =
for i in range(n+1):
if i == 0 or i == 1:
else:
return lst.pop()
fib_loop(5)
def fib_loop2(n):
a, b = 1, 1
for i in range(n+1):
if i == 0 or i == 1:
a, b = 1, 1
else:
a, b = b, a+b
return b
fib_loop2(5)
生成器：
def fib_gen():
i = 0
a, b = 1, 1
while true:
if i == 0 or i == 1:
yield 1
else:
a, b = b, a+b
yield b
i += 1
def fib(n):
gen = fib_gen()
for _ in range(n):
next(gen)
return next(gen)
fib(5)
2.寫乙個函式，實現對整數的排序，預設公升序排序，不能使用任何內建函式和第三方庫
fun1：
def sort_int(lst, asc=true):
sort_lst =
while lst:
i = none
min_int = none
for i in lst:
if min_int is none:
min_int = i
else:
if min_int > i:
min_int = i
lst.remove(min_int)
if asc:
else:
sort_lst.insert(0, min_int)
return sort_lst
fun2：
def sort(lst, reverse=false):
dst =
for n in lst:
for i, e in enumerate(dst):
if not reverse:
if n < e:
dst.insert(i, n)
break
else:
if n > e:
dst.insert(i, n)
break
else:
return dst
3.寫乙個函式，把整數轉化為羅馬數字，輸入為1到3999之間的任意數字
def int2rom(n):
if n < 1 or n > 3999:
print('input error.')
return 0
rom_lst = [' ',
'i', 'ii', 'iii', 'iv', 'v', 'vi', 'vii', 'viii', 'ix',
'x', 'xx', '***', 'xl', 'l', 'lx', 'lxx', 'l***', 'xc',
'c', 'cc', 'ccc', 'cd', 'd', 'dc', 'dcc', 'dccc', 'cm',
'm', 'mm', 'mmm']
rom = ''
bit = n % 10
ten = (n % 100) // 10
hun = (n % 1000) // 100
tho = n // 1000
if tho != 0:
rom += rom_lst[27 + tho]
if hun != 0:
rom += rom_lst[18 + hun]
if ten > 0:
rom += rom_lst[9 + ten]
if bit > 0:
rom += rom_lst[bit]
return rom
3.2寫乙個函式，把整數轉化為羅馬數字，輸出為1到3999之間的任意數字
def roman2int(src):
convert_map = {
'i':1,
'v':5,
'x':10,
'l':50,
'c':100,
'd':500,
'm':1000
# mdcccxcix
# xicxcccdm
# +10 -1 +100 -10 +100 +100 +100 +500 +1000
# 1899
roman = src.upper()[::-1]
prev = 0
lst =
for x in roman:
i = convert_map[x]
if i < prev:
else:
prev = i
return sum(lst)
roman2int('mdcccxcix')
4.寫乙個函式，求兩個字串的最長公共子串
例如輸入：i love python 和 python is a ****** language，輸出為python
def psubstring_long(x, y):
f =
max_len = 0
idx = 0
for i, a in enumerate(x):
for j, b in enumerate(y):
if a != b:
else:
if i == 0 or j == 0:
else:
if max_len < f[i][j]:
max_len = f[i][j]
# idx = max_len - (i+1)
idx = i + 1 - max_len
return x[idx: idx + max_len]
psubstring_long('babcdf', 'xxaabcaac')
5.動態規劃
總長n，step表示一次可選步數，總共多少種走法
ret = 0
tmp =
solution =
def recursion(n, tmp, step=3):
global ret
if n == 0:
solution.extend(tmp)
ret += 1
return 0
for i in range(1, step+1):
if n > i-1:
recursion(n-i, tmp, step)
tmp.pop(-1)
return ret
print(recursion(3, , 3), solution)
4 ['===', 1, 1, 1, '===', 1, 2, '===', 2, 1, '===', 3]
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