在数组中查找单个出现的元素

Question

I'm trying to find single occurrence of an element in an array. But it only shows for 1 element. Where is the logic going wrong?

function findSingle(array){
  var arrayCopy = array.slice(0);
  var x;
  var y = [];

  for (var i = 0; i < array.length; i++) {
    x = arrayCopy.splice(i, 1)
    if(arrayCopy.includes(array[i]) === false){
      console.log(array[i] + " is single")
    }
    arrayCopy = arrayCopy.concat(x)
   }
 }

findSingle([1, 3, 3, 6])

Answer 1

You can use a double Array.filter() to remove numbers that appear more than once:

 function findSingle(arr) { return arr.filter(i => arr.filter(j => i === j).length === 1) } const result = findSingle([1, 3, 3, 6, 8, 4, 6]) console.log(result) // [1, 8, 4]

Bear in mind that this will only work for Numbers and other primitives because of the way that Javascript evaluates equality and sameness .

Answer 2

I added some console.logs to see what was happening and the problem is that you're changing the order of elements in arrayCopy . And so 6 is never checked.

Checking for [ 1 ]
arrayCopy is [ 3, 3, 6 ]
1 is single
After adding to arrayCopy [ 3, 3, 6, 1 ]
Checking for [ 3 ]
arrayCopy is [ 3, 6, 1 ]
After adding to arrayCopy [ 3, 6, 1, 3 ]
Checking for [ 1 ]
arrayCopy is [ 3, 6, 3 ]
After adding to arrayCopy [ 3, 6, 3, 1 ]
Checking for [ 1 ]
arrayCopy is [ 3, 6, 3 ]
After adding to arrayCopy [ 3, 6, 3, 1 ]

You can probably use a frequency map to find the number of occurrences in each element and then filter keys that occur only once.

function findSingle(array){
  var freqs = {};
  array.forEach(n => {
    if (!(n in freqs)) freqs[n] = 1;
    else freqs[n] += 1;
  });
  return Object.keys(freqs).filter(k => freqs[k] === 1);
}

Answer 3

This could be done by creating an object that could map element and its occurrence. So here is a code

function findSingle(arr){
    var counts = {};
    var singles = [];
    for (var i = 0; i < arr.length; i++) {
        var num = arr[i];
        counts[num] = counts[num] ? counts[num] + 1 : 1;
    }
    for(var num in counts) {
        if(counts[num] == 1)
            singles.push(num);
    }
    return singles.map(x => Number(x));
}

The output of findSingle([1, 3, 3, 6]) will be

[1, 6]

Note This can work with string too but it may be numbers. For example, ["1", "3", "3", "6"]