Question

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

方法一：字符串分割

我们可以将字符串 $text$ 按照空格分割成字符串数组 $words$，然后遍历 $words$，如果 $words[i]$ 和 $words[i+1]$ 分别等于 $first$ 和 $second$，那么就将 $words[i+2]$ 添加到答案中。

遍历结束后，返回答案列表。

时间复杂度 $O(L)$，空间复杂度 $O(L)$，其中 $L$ 是 $text$ 的长度。

class Solution:
  def findOcurrences(self, text: str, first: str, second: str) -> List[str]:
    words = text.split()
    ans = []
    for i in range(len(words) - 2):
      a, b, c = words[i : i + 3]
      if a == first and b == second:
        ans.append(c)
    return ans

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]);
  }
}

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

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

function findOcurrences(text: string, first: string, second: string): string[] {
  const words = text.split(' ');
  const n = words.length;
  const ans: string[] = [];
  for (let i = 0; i < n - 2; i++) {
    if (words[i] === first && words[i + 1] === second) {
      ans.push(words[i + 2]);
    }
  }
  return ans;
}