Single 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 O(n).

Example 1:

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

Example 2:

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

Solution:

function singleNumber(nums) {
    for (let i = 0; i < nums.length; i++) {
        let n1 = nums[i];
        if (nums.indexOf(n1) === nums.lastIndexOf(n1)) {
            return n1;
        }
    }
};
console.log(singleNumber([2, 2, 1]));
console.log(singleNumber([4, 1, 2, 1, 2]));
console.log(singleNumber([1, 1, 1, 1, 1, 5]));

Solution 2:

function singleNumber2(nums) {
    return nums.reduce((sum, num) => sum ^ num);
}
console.log(singleNumber2([2, 2, 1]));
console.log(singleNumber2([4, 1, 2, 1, 2]));
console.log(singleNumber2([1, 1, 1, 1, 1, 5]));

• // ^ is XOR