Single Number III

Description:

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?

Code:

vector<int> singleNumber(vector<int>& nums) {
vector<int>result;
if (nums.size() < 2)
return result;
int xorAll = 0;
for (int i = 0; i < nums.size(); ++i)
xorAll^=nums[i];
unsigned int x = 1;
while ((xorAll & x) == 0)
{
x=x<<1;
}
int result1=0,result2=0;
for (int j = 0; j < nums.size(); ++j)
{
if ((nums[j]&x) == 0)
result1^=nums[j];
else
result2^=nums[j];
}
result.push_back(result1);
result.push_back(result2);
return result;
}

PS:

思路很清晰，但是写代码的是后老在细节出错，主要是两个判断条件要注意