![](/img/trans.png)
[英]How to find the smallest four digit number, whose digits dont repeat, but add up to a randomized number
[英]Smallest number whose digits are decreasing
我發現了一個實現函數的問題,該函數接受一個正整數n作為輸入並返回大於n的最小正整數,該正整數的位數在減小,而對於返回大於n的最小正整數的函數,其正整數又遞增。 我認為增加功能可以正常工作。 但是函數減少有什么錯誤呢? 對於減少的輸入(100),它返回11而不是110。
# the next integer whose digits are increasing.
def increasing(n):
asastring = str(n)
length = len(asastring)
if asastring == "9"*length:
return "1"*(length+1)
if length == 1:
return int(n)+1
if length >= 2:
firstcharacter = asastring[0]
secondcharacter = asastring[1]
if int(firstcharacter) > int(secondcharacter):
return int(str(firstcharacter)*length)
if firstcharacter == secondcharacter:
return firstcharacter+str(increasing(int(asastring[1:])))
if int(firstcharacter) < int(secondcharacter):
if secondcharacter == "9":
return str(int(firstcharacter)+1) * len(str(n))
return firstcharacter+str(increasing(int(asastring[1:])))
# the next integer whose digits are decreasing.
def decreasing(n):
asastring = str(n)
length = len(asastring)
# First the case where we need to add a digit.
if asastring == "9"*length:
return "1"+"0"*length
# Now we know that the next integer has the same number of digits as the original number.
if length == 1:
return int(n)+1
if length >= 2:
firstcharacter = asastring[0]
secondcharacter = asastring[1]
if int(firstcharacter) > int(secondcharacter):
endpart = str(((asastring[1:3])))
value = firstcharacter + str(decreasing(int(asastring[1:])))
return str(firstcharacter) + str(decreasing(int(asastring[1:])))
if int(firstcharacter) == int(secondcharacter):
return decreasing(firstcharacter+str(decreasing(int(asastring[1:]))))
if int(firstcharacter) < int(secondcharacter):
return str(int(firstcharacter)+1)+'0'*(length-1)
i=100
print(increasing(i))
print(decreasing(i))
您需要刪除在遞歸調用中完成的int類型轉換,因為int('00')將您的數字轉換為零(基本上刪除了所有開頭的零)並縮短了字符串的長度。 只需刪除該轉換即可。.其余代碼工作正常:
def decreasing(n):
asastring = str(n)
length = len(asastring)
# First the case where we need to add a digit.
if asastring == "9"*length:
return "1"+"0"*length
# Now we know that the next integer has the same number of digits as the original number.
if length == 1:
return int(n)+1
if length >= 2:
firstcharacter = asastring[0]
secondcharacter = asastring[1]
if int(firstcharacter) > int(secondcharacter):
return str(firstcharacter) + str(decreasing(asastring[1:]))
if int(firstcharacter) == int(secondcharacter):
return decreasing(firstcharacter+str(decreasing(asastring[1:])))
if int(firstcharacter) < int(secondcharacter):
return str(int(firstcharacter)+1)+'0'*(length-1)
有多個相互交織的問題需要解決。 這兩個功能的命名令人困惑。 如果我們遵循邏輯,則函數increasing()
應稱為nondecreasing()
,類似地,函數decreasing()
應稱為nonincreasing()
。 >(大於)和> =(大於或等於)之間的區別。
下一個困惑是這些函數接受並返回哪種類型? 如果我們檢查工作中的 increasing()
函數返回的內容,則會得到:
str return "1"*(length+1)
int return int(n)+1
int return int(str(firstcharacter)*length)
str return firstcharacter+str(increasing(int(asastring[1:])))
str return str(int(firstcharacter)+1) * len(str(n))
str return firstcharacter+str(increasing(int(asastring[1:])))
如果我們類似地查看increasing()
如何處理其自身的內部遞歸調用,以查看其認為接受和返回的內容,則會得到:
int -> int return firstcharacter+str(increasing(int(asastring[1:])))
int -> int return firstcharacter+str(increasing(int(asastring[1:])))
因此,這里嘗試了increasing()
,又稱為nondecreasing()
重做,試圖使其始終接受一個int
並返回一個int
:
def nondecreasing(n): # aka increasing()
as_string = str(n)
length = len(as_string)
if as_string == "9" * length:
return int("1" * (length + 1))
if length == 1:
return int(n) + 1
first_digit, second_digit, second_digit_onward = as_string[0], as_string[1], as_string[1:]
if first_digit > second_digit:
return int(first_digit * length)
if first_digit == second_digit:
return int(first_digit + str(nondecreasing(int(second_digit_onward))))
if as_string == first_digit + "9" * (length - 1):
return int(str(int(first_digit) + 1) * length)
return int(first_digit + str(nondecreasing(int(second_digit_onward))))
decreasing()
aka nonincreasing()
函數的問題更大。 它依靠接受int
或內部調用的str
來解決問題的能力。
討論這些問題,而不是讓其他程序員重新發現它們,是代碼注釋的全部內容。
我不認為上述問題會阻止nonincreasing()
始終返回 int
:
def nonincreasing(n): # aka decreasing()
as_string = str(n)
length = len(as_string)
if as_string == "9" * length:
return int("1" + "0" * length)
if length == 1:
return int(n) + 1
first_digit, second_digit, second_digit_onward = as_string[0], as_string[1], as_string[1:]
if first_digit > second_digit:
return int(first_digit + str(nonincreasing(second_digit_onward)))
if first_digit == second_digit:
remaining_digits = str(nonincreasing(second_digit_onward))
second_digit = remaining_digits[0]
n = first_digit + remaining_digits
if first_digit < second_digit:
return int(str(int(first_digit) + 1) + '0' * (length - 1))
return int(n)
修復此功能的關鍵是從倒數第二個if
子句中刪除return
語句,而是修復數據,然后讓它進入下一個if
子句,以查看結果是否需要修復。
我相信@ devender22對int()
強制轉換的見解是至關重要的,但是我不認為隨附的解決方案是有效的,因為它會生成大量錯誤結果(例如,從990到998到1000時它們都應該被簡單地撞掉) 1)。
為了檢查我的nonincreasing()
函數的所有情況,我使用完全不同的Python運算符編寫了一個效率較低,非遞歸的解決方案,沒有單獨的情況:
def nonincreasing(n):
def is_increasing(n):
string = str(n)
return any(map(lambda x, y: y > x, string, string[1:]))
while is_increasing(n + 1):
n += 1
return n + 1
然后確保兩個實現在其輸出上達成一致。
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.