最长的单词-JS

Question

I am writing a function that returns the longest string in the given array. If the array is empty, it should return an empty string (""). If the array contains no strings; it should return an empty string.

function longestWord(arr) {
 var filtered = arr.filter(function(el) { return typeof el == 'number' });
  if (filtered.length > 0) {
    return Math.min.apply(Math, filtered);
  } else {
    return 0;
  }
}

var output = longestWord([3, 'word', 5, 'up', 3, 1]);
console.log(output); // --> must be 'word'

Right now my codes doesnt pull the word instead it pulls out the number. Any idea what am I missing?

Answer 1

Let's walk through your code.

The first line of your longestWord function:

var filtered = arr.filter(function(el) { return typeof el == 'number' });

will filter the input array based on typeof el === 'number' , which will return an array containing only the elements of the input array which are type of === number .

Since the goal is to find the longest word, this should probably be changed to:

var filtered = arr.filter(function(el) { return typeof el === 'string' });

which will return an array of the strings in the input array.

Next, there's a check to see if the filtered array is empty. If the array is empty, you return 0 . Your instructions say that if the array is empty, or if the array contains no strings, it should return an empty string. So we should change this to:

return "";

If the array is not empty, or contains strings, Math.min.apply(Math, filtered) is returned. This statement would return the minimum value of an array, so probably not what you want. After all, the goal is to return the longest string.

To do this we can use a variety of methods, here's one:

filtered.reduce(function(a, b) { return a.length > b.length ? a : b })

This statement uses the reduce() method to step through the array and return the longest item.

Putting it all together we get:

 function longestWord(arr) { var filtered = arr.filter(function(el) { return typeof el === 'string' }); if (filtered.length > 0) { return filtered.reduce(function(a, b) { return a.length >= b.length ? a : b }); } else { return ""; } } console.log(longestWord([3, 'word', 5, 'up', 3, 'testing', 1])); console.log(longestWord([])); console.log(longestWord([1, 2, 3, 4, 5])) console.log(longestWord(['some', 'long', 'four', 'char', 'strs'])) 

Answer 2

I assume the filter meant to be for "string"

Once you have the filtered array, you can simply use reduce to get the longest word

 function longestWord(arr) { return Array.isArray(arr) ? arr .filter(el => typeof el == 'string' ) .reduce((a,b) => b.length > a.length ? b : a) : ''; } console.log(longestWord([3, 'word', 5, 'up', 3, 1, 'blah'])); 

Even shorter code

var longestWord = (arr) => Array.isArray(arr) ? arr.reduce((a,b) => (typeof b == 'string' && b.length > a.length) ? b : a, '') : '';

Answer 3

const longestWord = arr =>
  arr.reduce((result, str) =>
    typeof str == 'string' && result.length < str.length
      ? str 
      : result, '')

Answer 4

This isn't the most efficient code. For that, seek other answers, especially those that use reduce , however I do find it more readable and thus easier to maintain:

 function longestWord(arr) { var filtered = arr.filter(el => typeof el === 'string') .sort((a,b) => a.length > b.length ? -1 : 1 ); if (filtered.length) return filtered[0]; return null; } var output = longestWord([3, 'word', 5, 'up', 3, 1]); console.log(output); // --> must be 'word' 

But what to do when two strings are the same length?

Answer 5

As all other answers are using filter() , reduce() and all that new and fancy methods, I'm gonna answer with the do-it-yourself method (ie the old fashion way).

function longestWord(arr) {
    if ( ! ( typeof arr === 'array' ) ) {
        console.log("That's not an array!");
        return '';
    }

    var longest = ''; // By default, the longest word is the empty string

    // Linear search
    for ( var i = 0, length = arr.length; i < length; i++ ) {
        var el = arr[i];
        if ( typeof el === 'string' && el.length > longest.length ) {
            // Words STRICTLY GREATER than the last found word.
            // This way, only the first word (if two lengths match) will be considered.
            longest = el;
        }
    }

    return longest;
}

I know, this is probably NOT what you want since you're already using filtered() , but it works.