260. 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.

For example:

Given 

nums = [1, 2, 1, 3, 2, 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?

一路XOR下来可以找出两个数XOR以后的数，到这里思路就断了。。用的O(n)空间hash解决.

public int[] singleNumber(int[] nums)
{
int len=nums.length;
if(len<1)
return new int[]{};

HashSet<Integer> hashset=new HashSet<>(len);
for(int a:nums)
if(!hashset.contains(a))
hashset.add(a);
else {
hashset.remove(a);
}

int[] retarr=new int[hashset.size()];
Iterator<Integer> it=hashset.iterator();
int cnt=0;
while(it.hasNext())
retarr[cnt++]=it.next();

return retarr;
}

得到两个数的XOR以后的处理办法，摘自：
https://leetcode.com/discuss/52351/accepted-java-space-easy-solution-with-detail-explanations
Once again, we need to use XOR to solve this problem. But this time, we need to do it in two passes:

In the first pass, we XOR all elements in the array, and get the XOR of the two numbers we need to find. Note that since the two numbers are distinct, so there must be a set bit (that is, the bit with value '1')
in the XOR result. Find out an arbitrary set bit (for example, the rightmost set bit).

In the second pass, we divide all numbers into two groups, one with the aforementioned bit set, another with the aforementinoed bit unset. Two different numbers we need to find must fall into thte two distrinct
groups. XOR numbers in each group, we can find a number in either group.

public int[] singleNumber(int[] nums) {
// Pass 1 :
// Get the XOR of the two numbers we need to find
int diff = 0;
for (int num : nums) {
diff ^= num;
}
// Get its last set bit
diff &= -diff;

// Pass 2 :
int[] rets = {0, 0}; // this array stores the two numbers we will return
for (int num : nums)
{
if ((num & diff) == 0) // the bit is not set
{
rets[0] ^= num;
}
else // the bit is set
{
rets[1] ^= num;
}
}
return rets;
}