面试题38：字符串的排列（可以和集合的子集对比）

# -*- coding:utf-8 -*-
class Solution:
    def Permutation(self, ss):
        # write code here
        if len(ss) == 0:
            return []
        if len(ss) == 1:
            return [ss]
        #循环遍历每个字符，让每个字符打头，然后继续递归遍历后边的字符
        charlist = list(ss)
        result = []

        for i in range(len(charlist)):
            for j in self.Permutation(''.join(charlist[:i])+''.join(charlist[i+1:])):
                if charlist[i]+j not in result:
                    result.append(charlist[i]+j)
        return result