single number I

Given an array of integers, every element appears twice except for one. Find that single one.

object Solution {    def singleNumber(nums: Array[Int]): Int = {        return nums.reduce(_^_);      }  }

single number II

Given an array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

def singleNumber(nums: Array[Int]): Int = {    def loop(nums: Array[Int], len: Int, index: Int, a: Int, b: Int): Int = index match {      case `len` => b      case _ => loop(nums, len, index + 1, a ^ nums(index) & ~((b ^ nums(index)) & ~a), (b ^ nums(index)) & ~a)    }    loop(nums, nums.length , 0, 0, 0)  }

single number III

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

def singleNumber(nums: Array[Int]): Array[Int] = {    def loop(nums: Array[Int], len: Int, idx: Int, res: Int, dec: Int, a: Int, b: Int): Array[Int] = (idx, res) match {      case (`len`, 0) => Array(a ^ nums(idx), b)      case (`len`, _) => Array(a, b ^ nums(idx))      case (_, 0) => loop(nums, len, idx + 1, nums(idx + 1) & dec, dec, a ^ nums(idx), b)      case _ => loop(nums, len, idx + 1, nums(idx + 1) & dec, dec, a, b ^ nums(idx))    }    val diff = nums.reduce(_ ^ _)    val dec = diff & -diff    loop(nums, nums.length - 1, 0, nums(0) & dec, dec, 0, 0)  }