884 Uncommon Words from Two Sentences 两句话中的不常见单词
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:
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.
题目描述:
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 :
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。
思路:
将 A和 B加起来看作一个字符串, 按空格分开(split(" "))之后用 map记录单词和出现次数, 返回只出现一次的单词即可
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
class Solution
{
public:
vector uncommonFromSentences(string A, string B)
{
vector result;
A += " " + B;
map word_times;
split(A, result, " ");
for (auto word : result) word_times[word]++;
result.clear();
for (auto word : word_times) if (word.second < 2) result.push_back(word.first);
return result;
}
private:
int split(const string& str, vector& result, string separator = ",")
{
if (str.empty())
{
return 0;
}
string temp;
string::size_type pos_begin = str.find_first_not_of(separator);
string::size_type comma_pos = 0;
while (pos_begin != string::npos)
{
comma_pos = str.find(separator, pos_begin);
if (comma_pos != string::npos)
{
temp = str.substr(pos_begin, comma_pos - pos_begin);
pos_begin = comma_pos + separator.length();
}
else
{
temp = str.substr(pos_begin);
pos_begin = comma_pos;
}
if (!temp.empty())
{
result.push_back(temp);
/* string.clear()
作用: 将字符串的内容清空，让源字符串成为一个空字符串（长度为0个字符）*/
temp.clear();
}
}
return 0;
}
};
Java:
class Solution {
public String[] uncommonFromSentences(String A, String B) {
A += " " + B;
String words[] = A.split(" ");
Map word_times = new HashMap<>();
for (String word : words) word_times.put(word, word_times.getOrDefault(word, 0) + 1);
List result = new ArrayList<>();
for (Map.Entry entry : word_times.entrySet()) if (entry.getValue() < 2) result.add(entry.getKey());
return result.toArray(new String[result.size()]);
}
}
Python:
class Solution:
def uncommonFromSentences(self, A: str, B: str) -> List[str]:
return [word for word in (A + " " + B).split(" ") if (A + " " + B).split(" ").count(word) == 1]