Force deduction question 500, keyboard line, JS implementation

Here's an array of strings words , Returns only those that can be used in American keyboard Words printed with letters on the same line . The keyboard is shown in the figure below .

American keyboard in ：

The first line consists of the characters "qwertyuiop" form .
The second line consists of characters "asdfghjkl" form .
The third line consists of the characters "zxcvbnm" form .

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/keyboard-row
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Example 1：

Input ：words = ["Hello","Alaska","Dad","Peace"]
Output ：["Alaska","Dad"]
Example 2：

Input ：words = ["omk"]
Output ：[]
Example 3：

Input ：words = ["adsdf","sfd"]
Output ：["adsdf","sfd"]

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/keyboard-row
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

###  Their thinking

Here is the solution

On the keyboard 3 Rows correspond to each other 3 An array

2 Secondary cycle , Take out each word for the first time, turn them into lowercase and then into array

The first word corresponds to which array it belongs

Start the second cycle if there is a letter of the word that does not belong to the array

Just exit the loop , Don't add the word to the array

###  Code

```javascript

/**

 * @param {string[]} words

 * @return {string[]}

 */

var findWords = function(words) {

   let a=['q','w','e','r','t','y','u','i','o','p']

   let b=['a','s','d','f','g','h','j','k','l']

   let c=['z','x','c','v','b','n','m'],flag=false,k=[]

   for(let i=0;i<words.length;i++){

       let n=words[i].toLowerCase().split('')

       if(a.includes(n[0])){

           for(let j=1;j<n.length;j++){

               if(!a.includes(n[j])){

flag=true

               break

               }

               

           }

       }else if(b.includes(n[0])){

           for(let j=1;j<n.length;j++){

               if(!b.includes(n[j]))

               {

                   flag=true

               break

               }

           }

       }else{

           for(let j=1;j<n.length;j++){

               if(!c.includes(n[j]))

               {flag=true

               break}

           }

       }

       if(!flag)k.push(words[i])

       flag=false

   }

   return k

};

```