English:

We are given two sentences A and B. (A sentence is a string of space separated words. Each word consists only of lowercase letters.)
A word is uncommon if it appears exactly once in one of the sentences, and does not appear in the other sentence.
Return a list of all uncommon words.
You may return the list in any order.
Example 1:
Input: A = "this apple is sweet", B = "this apple is sour"
Output: ["sweet","sour"]
Example 2:
Input: A = "apple apple", B = "banana"
Output: ["banana"]
Note:

1. 0 <= A.length <= 200
2. 0 <= B.length <= 200
3. A and B both contain only spaces and lowercase letters.

中文：

给定两个句子 A 和 B 。 （句子是一串由空格分隔的单词。每个单词仅由小写字母组成。）
如果一个单词在其中一个句子中只出现一次，在另一个句子中却没有出现，那么这个单词就是不常见的。
返回所有不常用单词的列表。
您可以按任何顺序返回列表。
示例 1：
输入：A = "this apple is sweet", B = "this apple is sour"
输出：["sweet","sour"]
示例 2：
输入：A = "apple apple", B = "banana"
输出：["banana"]
提示：
0 <= A.length <= 200
0 <= B.length <= 200
A 和 B 都只包含空格和小写字母。

1.

开始我是将两句分开来处理。明显显得很拥挤，代码有点啰嗦。

class Solution(object):
    def uncommonFromSentences(self, A, B):
        """
        :type A: str
        :type B: str
        :rtype: List[str]
        """        
        word_a = A.split(' ')
        d_a = {k:0 for k in word_a}
        word_b = B.split(' ')
        d_b = {k:0 for k in word_b}
        for x in word_a:
            d_a[x] += 1
        for y in word_b:
            d_b[y] += 1
        for k in d_a.keys():
            if d_a[k]==1:
                if k not in result and k not in d_b:
                    result.append(k)
        for k in d_b.keys():
            if d_b[k]==1:
                if k not in result and k not in d_a:
                    result.append(k)
        return result

理解还是很好理解的，后面我就把两个句子放到一起去处理了。

这是分开处理的

2.

我把两个句子加到一起，然后在处理，但是后面发现变慢了。。。。。

class Solution(object):
    def uncommonFromSentences(self, A, B):
        """
        :type A: str
        :type B: str
        :rtype: List[str]
        """
        st = A+' '+ B
        a = st.split(' ')
        s = {k:0 for k in st.split(' ')}
        result = []
        for l in a:
            s[l] += 1
        for i in a:
            if s[i]==1:
                if i not in result:
                    result.append(i)
        return result

image.png

我觉得是因为在合并语句时的原因，因为+对两个字符串使用，会新创建新的字符串实例，所以会慢，而且内存变大。