【LeetCode】 1408. String Matching in an Array

Description

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].
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.

給一個字串陣列words。以任何排序回傳所有在words中，是其他任意字串的子字串的字串。
如果words[i]在右邊或左邊加入任意字元後可以變成words[j]，字串words[i]是words[j]的子字串。
限制:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 30
• words[i] 只包含小寫英文字母。
• 保證words[i]的獨特性(不會重複)。

Example:

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: []

Solution

• C++中提供了find()的功能，可以直接使用。
  • 如果有找到就代表參數是物件的子字串。
• 為了避免重複，使用unique()將最後答案移除重複的元素。

Code

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[i].find(words[j]) != std::string::npos)
                    ans.push_back(words[j]);
            }
        }
        sort( ans.begin(), ans.end() );
        ans.erase( unique( ans.begin(), ans.end() ), ans.end() );
        return ans;
    }
};

tags: LeetCode C++