【leetcode】1239. Maximum Length of a Concatenated String with Unique Characters

题目如下：

Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters.

Return the maximum possible length of s.

Example 1:
Input: arr = ["un","iq","ue"]
Output: 4
Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique".
Maximum length is 4.
Example 2:
Input: arr = ["cha","r","act","ers"]
Output: 6
Explanation: Possible solutions are "chaers" and "acters".
Example 3:
Input: arr = ["abcdefghijklmnopqrstuvwxyz"]
Output: 26
Constraints:

1 <= arr.length <= 16

1 <= arr[i].length <= 26

arr[i] contains only lower case English letters.

解题思路：首先过滤掉arr中有重复字符的元素，接下来就可以采用BFS的方法进行计算了。怎么判断两个字符串是否有相同字符呢？我的方法是把字符串转成二进制的方式，a对应2^0,c对应2^2,z对应2^25。这样的话，每一个字符串都可以计算出一个特征值，判断两个字符串是否有相同字符，只有判断两个特征值与操作的结果是否为0即可。

代码如下：

class Solution(object):
    def maxLength(self, arr):
        """
        :type arr: List[str]
        :rtype: int
        """
        arr_val = []
        for i in arr:
            val = 0
            if len(set(list(i))) != len(i):continue
            for j in i:
                val |= 2**(ord(j) - ord('a'))
            arr_val.append(val)
        #print arr_val
        queue = []
        if len(arr_val) == 0:return 0
        for (inx,val) in enumerate(arr_val):
            queue.append((inx,val))
        res = 0
        while len(queue) > 0:
            inx,val = queue.pop(0)
            end = True
            for i in range(inx+1,len(arr_val)):
                if val & arr_val[i] != 0:continue
                queue.append((i,val | arr_val[i]))
                end = False
            if end :
                res = max(res,bin(val).count('1'))
        return res