comments | difficulty | edit_url | rating | source | tags | |
---|---|---|---|---|---|---|
true |
简单 |
1274 |
第 140 场周赛 Q1 |
|
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text
不包含任何前缀或尾随空格。
我们可以将字符串
遍历结束后,返回答案列表。
时间复杂度
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;
}