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);
};


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    Deyu

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