在数组（JS）中查找第一个唯一的非七字串

Question

I'm trying to find the first unique string in array ignoring all other anagrams for example :

let words = ['nap', 'teachers', 'cheaters', 'pan', 'ear', 'era', 'hectares'];

output = nap teachers ear

or let words = ['rat', 'tar', 'art'];

output rat

I stuck here

let set = new Set();
let words = let words = ['nap', 'teachers', 'cheaters', 'pan', 'ear', 'era', 'hectares']


let map = new Map();
for (let word of words) {
    let sorted = word.toLowerCase().split('').sort().join(''); 
    map.set(sorted, word);
    set.add(sorted)
}
let arr = Array.from(set)


for (let inx = 0; inx < words.length; inx++) {
    arr.push('0')
}
console.log(arr);
console.log(words);

function isAnagram(s1, s2){
    return s1.split("").sort().join("") === s2.split("").sort().join("");
  }

let result = [];

  for (let i = 0; i < words.length; i++) {
       if ()
       result.push(words[i])

}
console.log(result);

Answer 1

Instead of a Map of string: string in which you only store the last anagram option, you can create a map of string: [ string ] in which you store all options.

Then, in the end, you can loop over all those entries and return the first element of each array:

 let set = new Set(); let words = ['nap', 'teachers', 'cheaters', 'pan', 'ear', 'era', 'hectares'] let map = new Map(); for (let word of words) { let sorted = word.toLowerCase().split('').sort().join(''); if (!map.has(sorted)) map.set(sorted, []); set.add(sorted); map.get(sorted).push(word); } let arr = Array .from(set) .map(k => map.get(k)[0]) console.log(arr); 

Answer 2

You need to check the current position of the array "words" and then check the others one, this means you need to have a loop inside a loop, so your code should look like this:

 let set = new Set(); let words = ['nap', 'teachers', 'cheaters', 'pan', 'ear', 'era', 'hectares'] let map = new Map(); for (let word of words) { let sorted = word.toLowerCase().split('').sort().join(''); map.set(sorted, word); set.add(sorted) } let arr = Array.from(set) for (let inx = 0; inx < words.length; inx++) { arr.push('0') } console.log(arr); console.log(words); function isAnagram(s1, s2){ console.log("checking: " + s1 + " and " + s2); return s1.split("").sort().join("") === s2.split("").sort().join(""); } let result = []; let anagram = false; for (let i = 0; i < words.length; i++) { for(let j=i+1; j<words.length; j++){ if (isAnagram(words[i], words[j])){ anagram = true; break; console.log("Is anagram!"); } } if(anagram) { //Before insert we will see if we have an anagram in the result array anagram = false; for(let z=0; z<result.length; z++) { //We will se if it exists in result if(isAnagram(words[i],result[z])) { anagram = true; break; } } if(!anagram) { result.push(words[i]); anagram = false; } } } console.log(result); 

Answer 3

Thanks to user3297291 I found even simpler variant

let words = ['nap', 'teachers', 'cheaters', 'pan', 'ear', 'era', 'hectares']
const set = new Set();
const unique = [];

words.forEach((word) => {

 const sorted = word.split('').sort().join('');

 if (set.has(sorted)) {
   return;
 }
   unique.push(word);
   set.add(sorted);
 });

 print(unique.join(' '));

Answer 4

Another implementation. :)

function aClean(arr) {
     var map = new Map();
     arr.map(item => {
         var key = item.toLowerCase().split("").sort().join("");
         map.has(key) ? null : map.set(key, item)
     });
    return Array.from(map.values());
}

Answer 5

This should do the job:

let words = ['nap', 'teachers', 'cheaters', 'pan', 'ear', 'era', 'hectares']

let map = new Map(
    words.reverse().map(word => {
        let sorted = word.toLowerCase().split('').sort().join(''); 
        return [sorted, word];
    })
);

console.log(Array.from(map.values())); // output: [ "teachers", "ear", "nap" ]

Using the characters from each word as a key. Exemple: for [teacher, cheaters, hectares] the key is "aceehrst"

and since the order is important for finding the first one, reversing the order of words will help to keep "teacher" as a value for the key "aceehrst" instead of "hectares".