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[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
我发现了一个实现函数的问题,该函数接受一个正整数n作为输入并返回大于n的最小正整数,该正整数的位数在减小,而对于返回大于n的最小正整数的函数,其正整数又递增。 我认为增加功能可以正常工作。 但是函数减少有什么错误呢? 对于减少的输入(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))
您需要删除在递归调用中完成的int类型转换,因为int('00')将您的数字转换为零(基本上删除了所有开头的零)并缩短了字符串的长度。 只需删除该转换即可。.其余代码工作正常:
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)
有多个相互交织的问题需要解决。 这两个功能的命名令人困惑。 如果我们遵循逻辑,则函数increasing()
应称为nondecreasing()
,类似地,函数decreasing()
应称为nonincreasing()
。 >(大于)和> =(大于或等于)之间的区别。
下一个困惑是这些函数接受并返回哪种类型? 如果我们检查工作中的 increasing()
函数返回的内容,则会得到:
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:])))
如果我们类似地查看increasing()
如何处理其自身的内部递归调用,以查看其认为接受和返回的内容,则会得到:
int -> int return firstcharacter+str(increasing(int(asastring[1:])))
int -> int return firstcharacter+str(increasing(int(asastring[1:])))
因此,这里尝试了increasing()
,又称为nondecreasing()
重做,试图使其始终接受一个int
并返回一个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()
函数的问题更大。 它依靠接受int
或内部调用的str
来解决问题的能力。
讨论这些问题,而不是让其他程序员重新发现它们,是代码注释的全部内容。
我不认为上述问题会阻止nonincreasing()
始终返回 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)
修复此功能的关键是从倒数第二个if
子句中删除return
语句,而是修复数据,然后让它进入下一个if
子句,以查看结果是否需要修复。
我相信@ devender22对int()
强制转换的见解是至关重要的,但是我不认为随附的解决方案是有效的,因为它会生成大量错误结果(例如,从990到998到1000时它们都应该被简单地撞掉) 1)。
为了检查我的nonincreasing()
函数的所有情况,我使用完全不同的Python运算符编写了一个效率较低,非递归的解决方案,没有单独的情况:
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
然后确保两个实现在其输出上达成一致。
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