1408. String Matching in an Array

Given an array of string words. Return all strings in words which is substring of another word in any order. 

String words[i] is substring of words[j], if can be obtained removing some characters to left and/or right side of words[j].

Example 1:

Input: words = ["mass","as","hero","superhero"]
Output: ["as","hero"]
Explanation: "as" is substring of "mass" and "hero" is substring of "superhero".
["hero","as"] is also a valid answer.

Example 2:

Input: words = ["leetcode","et","code"]
Output: ["et","code"]
Explanation: "et", "code" are substring of "leetcode".

Example 3:

Input: words = ["blue","green","bu"]
Output: []

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 30
• words[i] contains only lowercase English letters.
• It’s guaranteed that words[i] will be unique.

---

给定一个字符串数组，如果某个字符串是其他字符串的子串，则输出，找出所有这种字符串。

简单题，直接两遍for循环查找即可，代码如下：

class Solution {
public:
	vector<string> stringMatching(vector<string>& words) {
		vector<string> ans;
		for (int i = 0; i < words.size(); ++i) {
			for (int j = 0; j < words.size(); ++j) {
				if (i == j)continue;
				if (words[j].find(words[i]) != string::npos) {
					ans.push_back(words[i]);
					break;
				}
			}
		}
		return ans;
	}
};

本代码提交AC，用时8MS。