LeetCode:Single Number III

problem:

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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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Solution:Bit Manipulation

class Solution {
public:
vector<int> singleNumber(vector<int>& nums) {

int AxorB=0;
for(int i=0;i<nums.size();i++)
{
AxorB^=nums[i];
}

//取最后一个二进制位  根据区别位将vector中的数分为两个序列
int mask=AxorB&(~(AxorB-1));
int A=0,B=0;

for(int i=0;i<nums.size();i++)
{
if(mask&nums[i])
A^=nums[i];
else
B^=nums[i];

}
return vector<int>({A,B});
}
};