原文链接: https://leetcode-cn.com/problems/uncommon-words-from-two-sentences

英文原文

A sentence is a string of single-space separated words where 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.

Given two sentences s1 and s2, return a list of all the uncommon words. You may return the answer in any order.

 

Example 1:

Input: s1 = "this apple is sweet", s2 = "this apple is sour"
Output: ["sweet","sour"]

Example 2:

Input: s1 = "apple apple", s2 = "banana"
Output: ["banana"]

 

Constraints:

• 1 <= s1.length, s2.length <= 200
• s1 and s2 consist of lowercase English letters and spaces.
• s1 and s2 do not have leading or trailing spaces.
• All the words in s1 and s2 are separated by a single space.

中文题目

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。

如果某个单词在其中一个句子中恰好出现一次，在另一个句子中却 没有出现 ，那么这个单词就是 不常见的 。

给你两个 句子 s1 和 s2 ，返回所有 不常用单词 的列表。返回列表中单词可以按 任意顺序 组织。

 

示例 1：

输入：s1 = "this apple is sweet", s2 = "this apple is sour"
输出：["sweet","sour"]

示例 2：

输入：s1 = "apple apple", s2 = "banana"
输出：["banana"]

 

提示：

• 1 <= s1.length, s2.length <= 200
• s1 和 s2 由小写英文字母和空格组成
• s1 和 s2 都不含前导或尾随空格
• s1 和 s2 中的所有单词间均由单个空格分隔

通过代码

官方题解

方法：计数

思路和算法

每个不常见的单词总共只出现一次。我们可以统计每个单词的出现次数，然后返回恰好出现一次的单词。

[xehjDBgG-Java]
class Solution {
    public String[] uncommonFromSentences(String A, String B) {
        Map<String, Integer> count = new HashMap();
        for (String word: A.split(" "))
            count.put(word, count.getOrDefault(word, 0) + 1);
        for (String word: B.split(" "))
            count.put(word, count.getOrDefault(word, 0) + 1);

        List<String> ans = new LinkedList();
        for (String word: count.keySet())
            if (count.get(word) == 1)
                ans.add(word);

        return ans.toArray(new String[ans.size()]);
    }
}

[xehjDBgG-Python]
class Solution(object):
    def uncommonFromSentences(self, A, B):
        count = {}
        for word in A.split():
            count[word] = count.get(word, 0) + 1
        for word in B.split():
            count[word] = count.get(word, 0) + 1

        #Alternatively:
        #count = collections.Counter(A.split())
        #count += collections.Counter(B.split())

        return [word for word in count if count[word] == 1]

复杂度分析

• 时间复杂度：$O(M + N)$，其中 $M, N$ 分别是 A 和 B 的长度。

• 空间复杂度：$O(M + N)$，count 所用去的空间。

统计信息

通过次数	提交次数	AC比率
20337	30686	66.3%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言