题解 | #字符串的排列#

class Solution:
    def Permutation(self , str1: str):
        s = list(str1)
        output = []
        n = len(s)

        def func(first):

            if first == n:
                output.append(''.join(s))
                return

            used = set()
            for i in range(first, n):
                if s[i] in used: # 压入集合的应该是s[i]，而不是i
                    continue
                
                used.add(s[i])

                s[first], s[i] = s[i], s[first]
                func(first + 1)
                s[first], s[i] = s[i], s[first]

        func(0)
        return output