排列组合的题，有点难度哦

爱知

参考网上的算法：

#!/usr/bin/env python def perm(seq, l=None):         if l is None:                 l = len(seq)         for i in range(len(seq)):                 c  = seq[i:i+1]                 if l == 1:                         yield c                 else:                         rest = seq[:i] + seq[i+1:]                         for p in perm(rest, l-1):                                 yield c + p def main(seq):         a = perm(seq)         b = set(a)         print "There are %i sequences:" %len(b)         for i in b:                 print i if __name__ == '__main__':         main('aaab')

运行结果：
$ python tst.py
There are 4 sequences:
abaa
aaba
aaab
baaa

[ 本帖最后由 爱知 于 2009-3-22 20:30 编辑 ]

victorlee129

ls 太麻烦了，如果只要输出个数的话：

1. 
   
2. #!/usr/bin/python
   
3. fact=lambda x : x and reduce(lambda a,b:a*b,xrange(1,x+1) or 1;
   
4. l=raw_input().split();
   
5. t=[l.count(l[i]) for i in xrange(len(l)) if i=l.index(l[i])];
   
6. x=fact(len(l));
   
7. for i in t : x=x/fact(i);
   

复制代码

niexining

发霉题配发霉算法：

def perm(s):     if len(s)==1:         yield s     else:         for x in set(s):             i=s.find(x)             for j in perm(s[:i]+s[i+1:]):                 yield s[i]+j

大家都被Teebye弄头晕了吧

niexining

想了想，find改成index更通用一点，既能用于list, 也能用于str.

千年沉寂

def judge(s,k,i):     for j in range(k,i):         if s[i] == s[j]:             return True     return False def reperm(s,k,m):     if k==m:         print s     else:         for i in range(k,m+1):             if judge(s,k,i):                 continue             s[k],s[i]=s[i],s[k]             perm(s,k+1,m)             s[k],s[i]=s[i],s[k] s=['a','a','b'] reperm(s,0,len(s)-1)

结果：

['a', 'a', 'b'] ['a', 'b', 'a'] ['b', 'a', 'a']

lemoncookie

可以打印出所有可能的排列，重复的组合不会打印出来

其中有些实现的地方很ugly，见笑

#!/usr/bin/python import copy l = ['a', 'b', 'c', 'd'] # xx() is a recursive function # it returns a list of list which represents all possible assembles for given list # NOTE: there might be identical assembles in the returned list def xx(ll):     result = list()     if len(ll) == 1:         result = ll        else:         # tail is the last element of current list         # pre is the list of possible assembles for ll[:-1]         tail = ll[-1]         pre = xx(ll[:-1])         # insert tail to all possible positions in pre         # use copy.deepcopy to create a new list from pre everytime         # put all newly generated assembles into result         for pre_sub in pre:             for i in xrange(len(pre_sub)):                 tmp = list(copy.deepcopy(pre_sub))                 tmp.insert(i, tail)                 result.append(tmp)             tmp = list(copy.deepcopy(pre_sub))             tmp.append(tail)             result.append(tmp)     return result if __name__ == '__main__':     t = xx(l)     # convert the return value of xx() from list of list     # to list of strings     for i in xrange(len(t)):         t[i] = reduce(lambda x, y: x + y, t[i])     # eleminate identical results with set()     t = set(t)     print 'Number of possible assembles for list', l, 'is:', len(t)     for ele in t:         print ele     

结果如下：

Number of possible assembles for list ['a', 'b', 'c', 'd'] is: 24 abcd bacd dbac dcab cabd cadb cdba dacb cdab bdca dbca cbda adcb acbd dcba acdb bdac cbad bcda dabc abdc adbc badc bcad

雨夜流星

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