Function singleNumber

single_number_136/solution.go:16–22

NOTE: bitwise XOR arithmetic operator used to find single digit n XOR 0 = n n XOR n = 0 So, for [2,2,1]: a = 0 ^ 2 = 2 a = 2 ^ 2 = 0 a = 0 ^ 1 = 1 = answer [2,1,2] a = 0 ^ 2 = 2 a = 2 ^ 1 = 3 (because 00000010 XOR 000000001 -> 00000011 -> 3) a = 3 ^ 2 = 1 (because 00000011 XOR 00000010 -> 00000001

(nums []int)

// a = 2 ^ 1 = 3 (because 00000010 XOR 000000001 -> 00000011 -> 3)
// a = 3 ^ 2 = 1 (because 00000011 XOR 00000010 -> 00000001 -> 1)
func singleNumber(nums []int) int {
	a := 0
	for i := 0; i < len(nums); i++ {
		a ^= nums[i]
	}
	return a
}

func singleNumber2(nums []int) int {
	seen := make(map[int]bool)