def list_all(string):
new_list_set = []
if len(string) == 1:
return string
for i in range(len(string)):
sub_string = string[:]
sub_string.pop(i)
result_list = list_all(sub_string)
for result in result_list:
if result+string[i] not in new_list_set:
new_list_set.append(result+string[i])
return new_list_set
string = list("asa")
for string_list in list_all(string):
print(string_list)
将数组的全排列看成是通过递归实现的,则只要将每次去除元素string[i]后的子串递归即可
剑指offer 字符串的排列 Python实现
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转载自blog.csdn.net/z2539329562/article/details/80807443
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