leetcode1079

class Solution(object):
    def __init__(self):
        self.count = 0
    
    def dfs(self, tiles, used, visited, path):
        
        if len(path) > 0 and path not in visited:
            visited.add(path)
            self.count += 1

        for i in range(len(tiles)):
            if used[i]:
                continue
            used[i] = True                            
            self.dfs(tiles, used, visited, path + tiles[i])        
            used[i] = False
                

    def numTilePossibilities(self, tiles):
        """
        :type tiles: str
        :rtype: int
        """

        visited = set()
        used = [False] * len(tiles)        
        self.dfs(tiles, used, visited, "")            
        return self.count

回溯法，与全排列使用list不同，这里要去掉同样的字符，因此使用set进行标记。（注意25行visited的定义）

参考：https://leetcode.com/problems/letter-tile-possibilities/discuss/308486/Simple-python-solution-with-backtracking

另外一行样式：

class Solution:
    def numTilePossibilities(self, tiles: str) -> int:
        return sum(len(set(itertools.permutations(tiles, i))) for i in range(1, len(tiles) + 1))

这样的写法很帅，但是可读性差一些，把代码分开来看：

import itertools

class Solution:
    def numTilePossibilities(self, tiles: str) -> int:
        sums = 0
        for i in range(1, len(tiles) + 1):
            permu = itertools.permutations(tiles, i)
            s = set(permu)
            sums += len(s)
        return sums

使用itertools，生成字符串的排列序列，简单又方便，效率也高。就是其他语言没有这种东西，不具有通用性。

参考：https://leetcode.com/problems/letter-tile-possibilities/discuss/308518/Python-1-liner