题解 | #字符串的排列#

def back(path,track):
            if len(track) == l:
                tmp = ''.join(track[:])
                if tmp not in result:
                    result.append(tmp)
                return

            for i in range(len(path)):
                p = path[i]
                track.append(p)

                back(path[:i]+path[i+1:], track)

                track.pop()

回溯算法的例子
输入参数path是待选择的路，track是走过的路的记录
根据track判断是否加入result
然后选出path的一种
做选择****
返回原始path