作業說明：https://inst.eecs.berkeley.edu/~cs61a/sp21/hw/hw03/#required-questions
Q1: Num eights
Write a recursive function num_eights that takes a positive integer x and returns the number of times the digit 8 appears in x. Use recursion - the tests will fail if you use any assignment statements.
傳入正整數x，返回x中數字8的個數，用遞歸，不能用賦值語句。
思路：要求使用遞歸，不能用賦值語句，那么計數只能在return語句中進行，num_eights(x // 10) + 1.
代碼：
def num_eights(x):
"""Returns the number of times 8 appears as a digit of x.
>>> num_eights(3)
0
>>> num_eights(8)
1
>>> num_eights(88888888)
8
>>> num_eights(2638)
1
>>> num_eights(86380)
2
>>> num_eights(12345)
0
>>> from construct_check import check
>>> # ban all assignment statements
>>> check(HW_SOURCE_FILE, 'num_eights',
... ['Assign', 'AugAssign'])
True
"""
"*** YOUR CODE HERE ***"
if x == 0:
return 0
if x % 10 == 8:
return num_eights(x // 10) + 1
return num_eights(x // 10)
Q2: Ping-pong
同樣要求要用遞歸，不能用賦值語句。
代碼：
def pingpong(n):
"""Return the nth element of the ping-pong sequence.
>>> pingpong(8)
8
>>> pingpong(10)
6
>>> pingpong(15)
1
>>> pingpong(21)
-1
>>> pingpong(22)
-2
>>> pingpong(30)
-2
>>> pingpong(68)
0
>>> pingpong(69)
-1
>>> pingpong(80)
0
>>> pingpong(81)
1
>>> pingpong(82)
0
>>> pingpong(100)
-6
>>> from construct_check import check
>>> # ban assignment statements
>>> check(HW_SOURCE_FILE, 'pingpong', ['Assign', 'AugAssign'])
True
"""
"*** YOUR CODE HERE ***"
先考慮賦值語句+while循環的常規解法：
def pingpong(n):
i = 1
ppvalue = 0
flag = 1
while i <= n:
ppvalue += flag
if num_eights(i) or i % 8 == 0:
flag = - flag
i += 1
return ppvalue
再用輔助函數替代while循環中的flag，遞歸調用
def pingpong(n):
def flag(x):
if x == 1:
return 1
if num_eights(x) or x % 8 == 0:
return -flag(x-1)
return flag(x-1)
if n == 1:
return 1
return pingpong(n-1) + flag(n-1)
Q3: Missing Digits
Write the recursive function missing_digits that takes a number n that is sorted in increasing order (for example, 12289 is valid but 15362 and 98764 are not). It returns the number of missing digits in n. A missing digit is a number between the first and last digit of n of a that is not in n. Use recursion - the tests will fail if you use while or for loops.
找到逐位增長（不減）的數字中，首位和末位間缺少的數字。
先考慮最小單元 n < 10 的情況，再考慮 n < 100時（即2位數）的情況，最后遞歸。
def missing_digits(n):
"""Given a number a that is in sorted, increasing order,
return the number of missing digits in n. A missing digit is
a number between the first and last digit of a that is not in n.
>>> missing_digits(1248) # 3, 5, 6, 7
4
>>> missing_digits(19) # 2, 3, 4, 5, 6, 7, 8
7
>>> missing_digits(1122) # No missing numbers
0
>>> missing_digits(123456) # No missing numbers
0
>>> missing_digits(3558) # 4, 6, 7
3
>>> missing_digits(35578) # 4, 6
2
>>> missing_digits(12456) # 3
1
>>> missing_digits(16789) # 2, 3, 4, 5
4
>>> missing_digits(4) # No missing numbers between 4 and 4
0
>>> from construct_check import check
>>> # ban while or for loops
>>> check(HW_SOURCE_FILE, 'missing_digits', ['While', 'For'])
True
"""
"*** YOUR CODE HERE ***"
if n < 10:
return 0
if n < 100:
return n % 10 - n // 10 - 1 if n % 10 != n // 10 else n % 10 - n // 10
return missing_digits(n // 10) + missing_digits(n % 100)
Q4: Count coins
