leetcode1079

 1 class Solution(object):
 2     def __init__(self):
 3         self.count = 0
 4     
 5     def dfs(self, tiles, used, visited, path):
 6         
 7         if len(path) > 0 and path not in visited:
 8             visited.add(path)
 9             self.count += 1
10 
11         for i in range(len(tiles)):
12             if used[i]:
13                 continue
14             used[i] = True                            
15             self.dfs(tiles, used, visited, path + tiles[i])        
16             used[i] = False
17                 
18 
19     def numTilePossibilities(self, tiles):
20         """
21         :type tiles: str
22         :rtype: int
23         """
24 
25         visited = set()
26         used = [False] * len(tiles)        
27         self.dfs(tiles, used, visited, "")            
28         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：

1 class Solution:
2     def numTilePossibilities(self, tiles: str) -> int:
3         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：

 1 import itertools
 2 
 3 class Solution:
 4     def numTilePossibilities(self, tiles: str) -> int:
 5         sums = 0
 6         for i in range(1, len(tiles) + 1):
 7             permu = itertools.permutations(tiles, i)
 8             s = set(permu)
 9             sums += len(s)
10         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

posted on 2019-06-09 21:57 Sempron2800+ 阅读(285) 评论(0) 编辑收藏举报