题解 | #字符串的排列#

思路：

遍历当前字符，再遍历除当前字符其他字符的全排列，相加为结果

1、注意重复去除

2、除以递归出口为字符长度为1时，直接返回该字符

class Solution:
    def Permutation(self , str: str) -> List[str]:
        # write code here
        
        if len(str) == 0:
            return []

        return self.helper(str)
    
    def helper(self, ss):
        if len(ss) <= 1: # 重点是递归出口
            return ss
        
        result = []
        for i in range(len(ss)):
            s1 = ss[i]
            for s2 in self.helper(ss[:i]+ss[i+1:]):
                s = s1 + s2
                if s not in result:
                    result.append(s)
        return result