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1078. Bigram 分词

English Version

题目描述

给出第一个词 first 和第二个词 second，考虑在某些文本 text 中可能以 "first second third" 形式出现的情况，其中 second 紧随 first 出现，third 紧随 second 出现。

对于每种这样的情况，将第三个词 "third" 添加到答案中，并返回答案。

 

示例 1：

输入：text = "alice is a good girl she is a good student", first = "a", second = "good"
输出：["girl","student"]

示例 2：

输入：text = "we will we will rock you", first = "we", second = "will"
输出：["we","rock"]

 

提示：

• 1 <= text.length <= 1000
• text 由小写英文字母和空格组成
• text 中的所有单词之间都由 单个空格字符 分隔
• 1 <= first.length, second.length <= 10
• first 和 second 由小写英文字母组成

解法

将 text 按空格切分为 words 列表，然后遍历 words，判断是否满足 words[i] == first && words[i + 1] == second，若是，则将 words[i + 2] 添加至结果列表 ans 中。

最后返回 ans 即可。

Python3

class Solution:
    def findOcurrences(self, text: str, first: str, second: str) -> List[str]:
        words = text.split(' ')
        return [
            words[i + 2]
            for i in range(len(words) - 2)
            if words[i] == first and words[i + 1] == second
        ]

Java

class Solution {

    public String[] findOcurrences(String text, String first, String second) {
        String[] words = text.split(" ");
        List<String> ans = new ArrayList<>();
        for (int i = 0; i < words.length - 2; ++i) {
            if (first.equals(words[i]) && second.equals(words[i + 1])) {
                ans.add(words[i + 2]);
            }
        }
        return ans.toArray(new String[0]);
    }
}

C++

class Solution {
public:
    vector<string> findOcurrences(string text, string first, string second) {
        istringstream is(text);
        vector<string> words;
        string word;
        while (is >> word) words.push_back(word);
        vector<string> ans;
        for (int i = 0; i < words.size() - 2; ++i)
            if (words[i] == first && words[i + 1] == second)
                ans.push_back(words[i + 2]);
        return ans;
    }
};

Go

func findOcurrences(text string, first string, second string) []string {
	words := strings.Split(text, " ")
	var ans []string
	for i := 0; i < len(words)-2; i++ {
		if words[i] == first && words[i+1] == second {
			ans = append(ans, words[i+2])
		}
	}
	return ans
}

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