def add_dots(s):
r = [s[:i] + '.' + s[i:] for i in range(1, len(s))]
r += [j + '.' + s[i:] for i in range(1, len(s)) for j in add_dots(s[:i])]
r += [s[:i] + '.' + j for i in range(1, len(s)) for j in add_dots(s[i:])]
return set(r)
//效率灰常低,纯属玩玩。。
p.s. 针对"abcde"字符串的排列
某男的ruby(1.9.x)精简版:
p (?b..?e).inject([?a]){|a,q|a.product [q,?.+q]}.map &:join
来个5行简单版
def add_dots(s): r = [s[:i] + '.' + s[i:] for i in range(1, len(s))] r += [j + '.' + s[i:] for i in range(1, len(s)) for j in add_dots(s[:i])] r += [s[:i] + '.' + j for i in range(1, len(s)) for j in add_dots(s[i:])] return set(r)//效率灰常低,纯属玩玩。。
p.s. 针对"abcde"字符串的排列
某男的ruby(1.9.x)精简版:
p (?b..?e).inject([?a]){|a,q|a.product [q,?.+q]}.map &:join简单地说就是笛卡尔积,至于看不看得懂是另一回事了……(反正我没看太懂,ruby语法太抽象。。)
某男的C精简版:
#define z(a,b) printf(#a"%s",(x>>b)&1?".":""), main(x){z(a,3)z(b,2)z(c,1)z(d,0)puts("e");16-x&&main(x+1);}与hit9同学协力完成了个(易读易写的)
from itertools import product [''.join(i + j for i, j in zip('abcd', p)) + 'e' for p in product(['.', ''], repeat = 4)]