1.分别使用递归、循环和生成器求菲波那切数列
递归：
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)
循环：
def fib_loop(n):
lst = []
for i in range(n+1):
if i == 0 or i == 1:
lst.append(1)
else:
lst.append(lst[-1] + lst[-2])
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.写一个函数，实现对整数的排序，默认升序排序，不能使用任何内置函数和第三方库
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:
sort_lst.append(min_int)
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:
dst.append(n)
return dst
3.写一个函数，把整数转化为罗马数字，输入为1到3999之间的任意数字
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', 'XXX', 'XL', 'L', 'LX', 'LXX', 'LXXX', '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写一个函数，把整数转化为罗马数字，输出为1到3999之间的任意数字
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:
lst.append(-1 * i)
else:
lst.append(i)
prev = i
return sum(lst)
roman2int('MDCCCXCIX')
4.写一个函数，求两个字符串的最长公共子串
例如输入：I love Python 和 Python is a simple language，输出为Python
def psubstring_long(x, y):
f = []
max_len = 0
idx = 0
for i, a in enumerate(x):
f.append([])
for j, b in enumerate(y):
if a != b:
f[i].append(0)
else:
if i == 0 or j == 0:
f[i].append(1)
else:
f[i].append(f[i-1][j-1] + 1)
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.动态规划
总长n，step表示一次可选步数，总共多少种走法
ret = 0
tmp = []
solution = []
def recursion(n, tmp, step=3):
global ret
if n == 0:
solution.append('===')
solution.extend(tmp)
ret += 1
return 0
for i in range(1, step+1):
if n > i-1:
tmp.append(i)
recursion(n-i, tmp, step)
tmp.pop(-1)
return ret
print(recursion(3, [], 3), solution)
4 ['===', 1, 1, 1, '===', 1, 2, '===', 2, 1, '===', 3]
