【Leetcode】：Single Number III问题 in Go语言

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.

For example:

Given 

nums = [1, 2, 1, 3, 2
4000
, 5]

, return 

[3,
5]

.

Note:

The order of the result is not important. So in the above example, 

[5, 3]

 is
also correct.

Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?
题目的大意是：给定一个数组，数组中的数除了两个互不相同的数外，其余数两两成对出现，现在要找出这两个数
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这道题拿在手上没什么简单的思路，参考了别人的结果“http://www.jianshu.com/p/402622b8ad43”
简要说下解题的思路，首先对所有数进行异或运算，那么相同的就会抵消掉，得到的结果就是两个相异的数a和b的异或，这个结果当作diff
然后求 (-diff & diff) 这个式子能得到二进制中只有一个1的数，这个1的位置代表着在这个位置上，a和b是不同的，那么就可以按照这个不同，
将所有数分为两组，第一组是该位置上与a相同，第二组是该位置上与b相同。

func singleNumber(nums []int) []int {
diff := 0
for _, v := range nums {
diff = diff ^ v
}
//得到的diff表示两个不同的数的按位异或的结果
diff = -diff & diff
result := make([]int, 2)
for _, v := range nums {
if diff & v == 0 {
result[0] = result[0] ^ v
} else {
result[1] = result[1] ^ v
}
}
return result
}