作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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目录

• * 题目描述
  • 题目大意
  • 解题方法
  • * 字符串分割遍历
  • 日期

题目地址：https://leetcode.com/problems/occurrences-after-bigram/

Given words first and second, consider occurrences in some text of the form "first second third", where second comes immediately after first, and third comes immediately after second.

For each such occurrence, add "third" to the answer, and return the answer.

Example 1:

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

Example 2:

Input: text = "we will we will rock you", first = "we", second = "will"
Output: ["we","rock"]

Note:

1. 1 <= text.length <= 1000
2. text consists of space separated words, where each word consists of lowercase English letters.
3. 1 <= first.length, second.length <= 10
4. first and second consist of lowercase English letters.

找出在一个长字符串text中，符合"first second third"模式的third。其中first和second是输入的。

字符串分割遍历

对字符串进行分割，然后循环遍历，每次判断三个单词，看前两个是不是first和second，如果满足的话那么把第三个放到结果中即可。

时间复杂度是O(N)。

Python代码如下：

class Solution(object):
    def findOcurrences(self, text, first, second):
        """
        :type text: str
        :type first: str
        :type second: str
        :rtype: List[str]
        """
        words = text.split()
        res = []
        L = len(words)
        for i in range(L - 2):
            f, s, t = words[i], words[i + 1], words[i + 2]
            if f == first and s == second:
                res.append(t)
        return res

C++代码的难点是字符串操作。C++竟然没有split()函数，这也是大家吐槽太多的点。一个可以替换的方案是使用stringstream类，该类会模拟字符串读入操作，即实现字符串按空格分割。初始化三个字符串，分别是f,s,t，读入到第三个里，前两个保留的是前两次读入的结果。这样进行判断是比较优雅的操作。

C++代码如下：

class Solution {
public:
    vector<string> findOcurrences(string text, string first, string second) {
        stringstream ss(text);
        vector<string> res;
        string f, s, t;
        while (ss >> t) {
            if (f == first && s == second)
                res.push_back(t);
            f = s;
            s = t;
        }
        return res;
    }
};

参考资料：
https://leetcode.com/problems/occurrences-after-bigram/discuss/308385/C%2B%2B-stringstream

2019 年 6 月 9 日 —— 简单的题没有难度，需要挑战有难度的才行