0

这个函数在函数内部有一个函数。该程序创建一个包含所有可能的字符串组合的列表。我在遵循内部函数时遇到了麻烦，它是递归的。有人可以帮帮我吗？

def combine0(s):
    out = []
    length = len(s)
    def loc(start):
        for i in range(start, length):
            out.append(s[i])
            print (''.join(out))
            if (i < length-1):
                loc(i+1)
            del out[-1]   
    loc(0)

combine0('mary')

生产

m
ma
mar
mary
may
mr
mry
my
a
ar
ary
ay
r
ry
y
4

2 回答 2

1

是的，那是一个调用自身的函数，名字叫recursive call。以下是此类代码的一些说明。

更正后的代码是：

def combine0(s):
    out = []
    length = len(s)
    def loc(start):
        for i in range(start, length):
            out.append(s[i])
            print (''.join(out))
            if (i < length-1):
                loc(i+1)
            del out[-1]
    loc(0)

combine0("abc")

“abc”的输出：

a ab abc ac b bc c

于 2013-09-26T13:37:48.117 回答
0

当使用参数调用时'Mary'，这就是combine0：

s = 'Mary'
out = []
length = len(s) # in this case length=4

# for i in range(0, 4):
# i = 0
out.append(s[0])    # s[0] = 'M', out = ['M']
print(''.join(out)) # print('M')
# if i < length - 1: => if 0 < 3:
# loc(i + 1) => loc(1)
# for j in range(1, 4):
# j = 1
out.append(s[1])    # s[1] = 'a', out = ['M', 'a']
print(''.join(out)) # print('Ma')
# if j < length - 1: => if 1 < 3:
# loc(j + 1) => loc(2)
# for k in range(2, 4):
# k = 2
out.append(s[2])    # s[2] = 'r', out = ['M', 'a', 'r']
print(''.join(out)) # print ('Mar')
# if k < length - 1: => if 2 < 3:
# loc(k + 1) => loc(3)
# for l in range(3, 4):
# l = 3
out.append(s[3])    # s[3] = 'y', out = ['M', 'a', 'r', 'y']
print(''.join(out)) # print('Mary')
# if l < length - 1: => if 3 < 3:
# condition fails -> do not recurse
del out[-1] # out = ['M', 'a', 'r']
# end of for l in range(3, 4)
del out[-1] # out = ['M', 'a']
# k = 3, now in for k in range(2, 4):
out.append(s[3])    # s[3] = 'y', out = ['M', 'a', 'y']
print(''.join(out)) # print('May')
# if k < length - 1: => if 3 < 3:
# condition fails -> do not recurse
del out[-1] # out = ['M', 'a']
# end of for k in range(2, 4)
del out[-1] # out = ['M']
# j = 2, now in for j in range (1, 4):
out.append(s[2])    # s[2] = 'r', out = ['M', 'r']
print(''.join(out)) # print('Mr')
# if j < length - 1: => if 2 < 3:
# loc(j + 1) => loc(3)
# for m in range(3, 4)
# m = 3
out.append(s[3])    # s[3] = 'y', out = ['M', 'r', 'y']
print(''.join(out)) # print('Mry')
# if m < length - 1: => if 3 < 3:
# condition fails -> do not recurse
del out[-1] # out = ['M', 'r']
# end of for m in range(3, 4)
del out[-1] # out = ['M']
# j = 3, now in for j in range (1, 4):
out.append(s[3])    # s[3] = 'y', out = ['M', 'y']
print(''.join(out)) # print('My')
# if j < length - 1: => if 3 < 3:
# condition fails -> do not recurse
del out[-1] # out = ['M']
# end of for j in range(1, 4)
del out[-1] # out = []
# i = 1

#... you get the rest
于 2013-09-26T15:17:53.317 回答