singleNumber.js

// /**
//  * @param {number[]} nums
//  * @return {number}
//  */

// Given a non - empty array of integers, every element appears twice except for one.Find that single one.

//     Note:

// Your algorithm should have a linear runtime complexity.Could you implement it without using extra memory ?

//     Example 1:

// Input: [2, 2, 1]
// Output: 1

// Example 2:

// Input: [4, 1, 2, 1, 2]
// Output: 4

// var singleNumber = function(nums) {
//   let i = 0;

//   nums.sort((a, b) => a - b);

//   for (i; i < nums.length; i++) {
//     if (nums[i - 1] != nums[i] && nums[i] != nums[i + 1]) {
//       return nums[i];
//     }
//   }
// };

var singleNumber = function(nums) {
  let count = new Map();
  for (num of nums) {
    if (count.get(num)) {
      count.set(num, count.get(num) + 1);
    } else {
      count.set(num, 1);
    }
  }

  let singleNum;

  count.forEach((value, key) => {
    if (value == 1) {
      singleNum = key;
    }
  });

  return singleNum;
};
console.log(singleNumber([4, 1, 2, 1, 2]));