题目地址：https://leetcode.com/problems/uncommon-words-from-two-sentences/description/
题目描述
Weare given two sentences A and B. (A sentence is a string of space separated words. Each word consists only of lowercase letters.)
Aword 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.
Youmay 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:
- 0 <= A.length <= 200
- 0 <= B.length <= 200
- A and B both contain only spaces and lowercase letters.
题目大意
如果一个词在一句话中只出现了一次，在另外一句话中没出现，那么这个词是不同的词。找出两句话中所有不同的词。
解题方法
字典统计
统计一下两句话单词的set，找出两个set的不同词，然后再判断这个词是否只出现了1次，如果只出现了1次，即为题目所求。
注意，要先找不同词，然后再判断是否出现1次。如这个测试用例：
Input:
"s z z z s"
"s z ejt"
Output:
["ejt","s","z"]
Expected:
["ejt"]
不可以先去重组成set，然后再求。。也就是说，只要一个词在一句话中出现的次数超过了1次，那么一定会被排除掉。
另外注意，这其实是求下面这个图的A+B部分。在python3中的Counter.keys是个set，可以直接做交并补操作。
代码如下：
class Solution:
def uncommonFromSentences(self, A, B):
"""
:type A: str
:type B: str
:rtype: List[str]
"""
count_A = collections.Counter(A.split(' '))
count_B = collections.Counter(B.split(' '))
words = list((count_A.keys() | count_B.keys()) - (count_A.keys() & count_B.keys()))
ans = []
for word in words:
if count_A[word] == 1 or count_B[word] == 1:
ans.append(word)
return ans
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