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 p(self,num, used, path, res):
        if len(path) == len(num):
            res.append(path[:])
            return
        for i in range(len(num)):
            
            if i>0 and num[i-1] == num[i] and used[i-1] == 0:
                continue
            if used[i] == 1:
                continue
            path.append(num[i])
            used[i] = 1
            self.p(num, used, path, res)
            path.pop()
            used[i] = 0
    def Permutation(self , str: str) -> List[str]:
        s = list(str)
        s.sort()
        str = ''.join(s)
        # write code here
        res = []
        a = []
        self.p(str, [0]*len(str), [],res)
        for i in res:
            a.append(''.join(i))
        return a