题解 | 字符串的排列

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param str string字符串 
# @return string字符串一维数组
#
class Solution:
    def Permutation(self , s: str) -> List[str]:
        str_list = [c for i,c in enumerate(s)]
        res_list = []
        path = []
        is_used = [False] * len(str_list)
        def backtrack():
            if len(path) == len(str_list):
                res_list.append(path.copy())
                return
            for i in range(len(str_list)):
                if is_used[i] == True:
                    continue
                path.append(str_list[i])
                is_used[i] = True
                backtrack()
                path.pop()
                is_used[i] = False
        
        backtrack()
        unique_list = [list(t) for t in set(tuple(lst) for lst in res_list)]
        result_list = [''.join(lst) for lst in unique_list]
        return sorted(result_list)