https://leetcode-cn.com/problems/single-number/
位运算法则:
a^0 == a
a^a == 0
a^b^b == a^0 == a
a^b == b^a
令 target 为 0, target^a^a^b^b 还是 0, target^x 为 x, 那么只出现一次的数最后被印在 target 上了.
Golang
func singleNumber(nums []int) int {
target := 0
for i := 0; i < len(nums); i++ {
target = target ^ nums[i]
}
return target
}