我编写下面的代码，根据0到9999范围内的5个数字的输入得出数字X.这5个数字每个都有两个数字，X也有，但它们的位置是错误的（例如一个数字）在千位中给出了数千位的X）。
示例输入：
6087,5173,1358,3825,2531
对于此输入，预期输出为：
8712
下面的代码计算X，但它是强力的。如何减少时间复杂度？
m = ('6087','5173','1358','3825','2531')
x = ['']*5
for i in range(0,5):
a=int(m[i])/1000
b=int(m[i])/100-10*a
c=int(m[i])/10-100*a-10*b
d=int(m[i])%10
x[i]=[a,b,c,d]
for num in range(0,10000):
thousand=num/1000
hundred=num/100-10*thousand
decade=num/10-100*thousand-10*hundred
digits=num%10
count=0
for j in range(0,5):
if (thousand==x[j][1] or thousand==x[j][2] or thousand==x[j][3]) and (hundred==x[j][0] or hundred==x[j][2] or hundred==x[j][3]) and (decade!=x[j][0] and decade!=x[j][1] and decade!=x[j][2] and decade!=x[j][3]) and (digits!=x[j][0] and digits!=x[j][1] and decade!=x[j][2] and decade!=x[j][3]):
count+=1
elif (thousand==x[j][1] or thousand==x[j][2] or thousand==x[j][3]) and (decade==x[j][0] or decade==x[j][1] or decade==x[j][3]) and (hundred!=x[j][0] and hundred!=x[j][1] and hundred!=x[j][2] and hundred!=x[j][3]) and (digits!=x[j][0] and digits!=x[j][1] and decade!=x[j][2] and decade!=x[j][3]):
count+=1
elif (thousand==x[j][1] or thousand==x[j][2] or thousand==x[j][3]) and (digits==x[j][0] or digits==x[j][1] or digits==x[j][2]) and (hundred!=x[j][0] and hundred!=x[j][1] and hundred!=x[j][2] and hundred!=x[j][3]) and (decade!=x[j][0] and decade!=x[j][1] and decade!=x[j][2] and decade!=x[j][3]):
count+=1
elif (hundred==x[j][0] or hundred==x[j][2] or hundred==x[j][3]) and (decade==x[j][0] or decade==x[j][1] or decade==x[j][3]) and (thousand!=x[j][0] and thousand!=x[j][1] and thousand!=x[j][2] and thousand!=x[j][3]) and (digits!=x[j][0] and digits!=x[j][1] and decade!=x[j][2] and decade!=x[j][3]):
count+=1
elif (hundred==x[j][0] or hundred==x[j][2] or hundred==x[j][3]) and (digits==x[j][0] or digits==x[j][1] or digits==x[j][2]) and (thousand!=x[j][0] and thousand!=x[j][1] and thousand!=x[j][2] and thousand!=x[j][3]) and (decade!=x[j][0] and decade!=x[j][1] and decade!=x[j][2] and decade!=x[j][3]):
count+=1
elif (decade==x[j][0] or decade==x[j][1] or decade==x[j][3]) and (digits==x[j][0] or digits==x[j][1] or digits==x[j][2]) and (thousand!=x[j][0] and thousand!=x[j][1] and thousand!=x[j][2] and thousand!=x[j][3]) and (hundred!=x[j][0] and hundred!=x[j][1] and hundred!=x[j][2] and hundred!=x[j][3]):
count+=1
if count == 5:
print num
答案 0 :(得分：0)
您可以使用它作为替代方法，执行递归搜索，并消除可能性，因为它选择的数字对可能是正确的（但错位）。如果它发现通过处理更多输入找不到解决方案，它会回溯并尝试下一对数字，等等。
一旦输入被处理到最深层次的递归，就会知道4个数字，并且另一个递归搜索会将这4个数字分配给仍然可能的位置。如果成功，函数会立即退出递归，返回找到的值。
def findMatch(guesses):
def assign(pos, digits, result):
if len(digits) == 0: return True
digit = digits[0]
for j in pos[digit]:
if result[j] == None:
result[j] = digit
if assign(pos, digits[1:], result):
return ''.join(map(str, result))
result[j] = None
return False
def recurse(guesses, pos, must):
if len(must) > 4: return False
if len(guesses) == 0:
if len(must) < 4: return False # all digits in solution must be unique
return assign(pos, list(must), [None, None, None, None])
digits = list(map(int, guesses[0]))
permutations = ((0,1),(2,3),(0,2),(1,3),(0,3),(1,2))
for i, (j, k) in enumerate(permutations):
(l, m) = permutations[i ^ 1]
[dig1, dig2, dig3, dig4] = [digits[j], digits[k], digits[l], digits[m]]
if (len(pos[dig1]) > int(j in pos[dig1]) and
len(pos[dig2]) > int(k in pos[dig2]) and
dig3 not in must and dig4 not in must):
saved = [pos[dig1], pos[dig2], pos[dig3], pos[dig4]]
# remove possibilities
[pos[dig1], pos[dig2], pos[dig3], pos[dig4]] = [
pos[dig1] - set([j]), pos[dig2] - set([k]), set(), set()]
found = recurse(guesses[1:], pos, must | set((dig1,dig2)))
if found:
return found
# backtrack
[pos[dig1], pos[dig2], pos[dig3], pos[dig4]] = saved
return False
return recurse(guesses, [set([0,1,2,3]) for i in range(0,10)], set())
guesses = ['6087','5173','1358','3825','2531']
print (findMatch(guesses))
这将比您实施的强力方法更快地找到解决方案。此代码假定解决方案必须由不同的数字组成 - 因此没有一个数字出现两次。