[Codewars/JS] Find the odd int

Given an array of integers, find the one that appears an odd number of times.

There will always be only one integer that appears an odd number of times.

Examples

[7] should return 7, because it occurs 1 time (which is odd).
[0] should return 0, because it occurs 1 time (which is odd).
[1,1,2] should return 2, because it occurs 1 time (which is odd).
[0,1,0,1,0] should return 0, because it occurs 3 times (which is odd).
[1,2,2,3,3,3,4,3,3,3,2,2,1] should return 4, because it appears 1 time (which is odd).

 

 

 

<Code>

function findOdd(A) {
  let count = 0;
  let result = 0;

  arr = A.sort(); //배열 정렬
  //console.log(arr); 
  for (let i = 0; i < arr.length; i++) { 
    if (arr[i] === arr[i + 1]) {// arr[i]와 arr[i+1]이 같을 때 까지 count ++
      count++;
    } else {
      count++; 

      if (count % 2 !== 0) { //만약 다르면 누적된 count값이 2의 배수가 아ㅣㄴ면 
        result = arr[i]; //그 값을 return
        break;
      }
    }
  }
  //console.log(result);
  return result;
}

findOdd([1, 1, 2, -2, 5, 2, 4, 4, -1, -2, 5]);