884. Uncommon Words from Two Sentences*

https://leetcode.com/problems/uncommon-words-from-two-sentences/

题目描述

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.

C++ 实现 1

使用哈希表存储 A 和 B 中每个 word 的个数, 取其中个数为 1 的 word. 注意:

A word is uncommon if it appears exactly once in one of the sentences, and does not appear in the other sentence.

class Solution {
public:
    vector<string> uncommonFromSentences(string A, string B) {
        unordered_map<string, int> record;
        stringstream ss1(A), ss2(B);
        string word;
        while (ss1 >> word) record[word] ++;
        while (ss2 >> word) record[word] ++;
        vector<string> res;
        for (auto &p : record)
            if (p.second == 1)
                res.push_back(p.first);
        return res;
    }
};

C++ 实现 2

class Solution {
public:
    vector<string> uncommonFromSentences(string A, string B) {
        unordered_set<string> setA, setA1, setB, setB1;
        stringstream ss1(A), ss2(B);
        string word;
        while (ss1 >> word) {
            if (!setA.count(word))
                setA.insert(word);
            else
                setA1.insert(word);
        }
        while (ss2 >> word) {
            if (!setB.count(word))
                setB.insert(word);
            else
                setB1.insert(word);
        }
            
        vector<string> res;
        for (auto &p : setA)
            if (!setB.count(p) && !setA1.count(p))
                res.push_back(p);
        for (auto &p : setB)
            if (!setA.count(p) && !setB1.count(p))
                res.push_back(p);
        return res;
    }
};