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 <= 1000text 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 <= 10first 和 second 由小写英文字母组成

解决方案

方法一：遍历

我们将文本 text 按空格分割成单词数组 words，然后遍历 words 数组，如果一个单词的前两个单词分别按顺序等于 first 和 second，则该单词符合第三个单词 third 的定义，将其加入结果中。

代码

C

char ** findOcurrences(char * text, char * first, char * second, int* returnSize){    int s = 0, e = 0, len = strlen(text);    char **words = (char **)malloc(sizeof(char *) * len);    memset(words, 0, sizeof(char *) * len);    int nwords = 0;    while (true) {        while (s < len && text[s] == ' ') {            s++;        }        if (s >= len) {            break;        }        e = s + 1;        while (e < len && text[e] != ' ') {            e++;        }        words[nwords++] = strndup(text + s, e - s);        s = e + 1;    }    char **ret = (char **)malloc(sizeof(char *) * (nwords + 1));    memset(ret, 0, sizeof(char *) * (nwords + 1));    int nret = 0;    for (int i = 2; i < nwords; i++) {        if (strcmp(words[i - 2], first) == 0 && strcmp(words[i - 1], second) == 0) {            ret[nret++] = strdup(words[i]);        }    }    for (int i = 0; i < nwords; i++) {        free(words[i]);    }    free(words);    *returnSize = nret;    return ret;}

C++

class Solution {public:    vector<string> findOcurrences(string text, string first, string second) {        vector<string> words;        int s = 0, e = 0, len = text.length();        while (true) {            while (s < len && text[s] == ' ') {                s++;            }            if (s >= len) {                break;            }            e = s + 1;            while (e < len && text[e] != ' ') {                e++;            }            words.push_back(text.substr(s, e - s));            s = e + 1;        }        vector<string> ret;        for (int i = 2; i < words.size(); i++) {            if (words[i - 2] == first && words[i - 1] == second) {                ret.push_back(words[i]);            }        }        return ret;    }};

Java

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

C#

public class Solution {    public string[] FindOcurrences(string text, string first, string second) {        string[] words = text.Split(" ");        IList<string> list = new List<string>();        for (int i = 2; i < words.Length; i++) {            if (words[i - 2].Equals(first) && words[i - 1].Equals(second)) {                list.Add(words[i]);            }        }        int size = list.Count;        string[] ret = new string[size];        for (int i = 0; i < size; i++) {            ret[i] = list[i];        }        return ret;    }}

Python3

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

Golang

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

JavaScript

var findOcurrences = function(text, first, second) {    const words = text.split(" ");    const list = [];    for (let i = 2; i < words.length; i++) {        if (words[i - 2] === first && words[i - 1] === second) {            list.push(words[i]);        }    }    const size = list.length;    const ret = Array(size).fill('');    for (let i = 0; i < size; i++) {        ret[i] = list[i];    }    return ret;};

复杂度分析

时间复杂度：O(N)，其中 N 为 text 的长度。分割 text 需要 O(N)，words 每个元素最多两次，需要 O(N)，所以总的时间复杂度为 O(N)。空间复杂度：O(N)。需要 O(N) 的空间来保存 words 数组。

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