作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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目录

• * 题目描述
  • 题目大意
  • 解题方法
  • * 字典统计
  • 日期

题目地址：https://leetcode.com/problems/uncommon-words-from-two-sentences/description/

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:

• 0 <= A.length <= 200
• 0 <= B.length <= 200
• A and B both contain only spaces and lowercase letters.

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。

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

2018 年 8 月 16 日 —— 一个月不写题，竟然啥都不会了。。加油！
2018 年 11 月 8 日 —— 项目进展缓慢