[ LeetCode ] [ JavaScript ] - 1408. String Matching in an Array 難度 ★ @ 西瓜是肥貓

先看看題目：

1408. String Matching in an Array

Easy

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.

先附上通關證明：

截圖 2020-06-13 下午6.14.53

我的想法：這一題就是說找出陣列裡面短字串有在另一個長字串之中。

其實用include去做就會很簡單。

就兩兩比對，在做長字串是否include短字串就好。

符合條件後就用個key value的物件塞進去，最後再把keys陣列提出就好。

/**
 * @param {string[]} words
 * @return {string[]}
 */
var stringMatching = function(words) {
    var result = {};
    for (var i = 0;i<words.length;i++) {
        for (var j = i+1 ; j<words.length;j++){
            if(i==j)continue;
            w1 = words[i] 
            w2 = words[j] 
            if(w1.length>w2.length){
                if(w1.includes(w2))
                result[w2] = true
                
            }else{
                if(w2.includes(w1))
                result[w1] = true
                
            }
        }
    }
    return Object.keys(result);
};