## Thursday, August 20, 2015

### LeetCode OJ - Single Number III

Problem:

Solution:

The XOR trick we used in Single Number can still be used with a twist. If we simply compute the overall XOR, it is the XOR of the two unknown single number.

A bit of the XOR of two number is 1 only if that bit differ for that two numbers. Therefore, if we (conceptually) separate the set of number by checking that bit, we can be sure that one of the number is on one side. So the remaining is easy. Just find a mask bit that is 1 on the overall XOR, compute the XOR of the subset with that mask on, that would be one of the number, the another can be found by using the overall XOR again.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/single-number-iii/

#include "LEET_SINGLE_NUMBER_III.h"
#include <map>
#include <iostream>
#include <vector>

using namespace std;

namespace _LEET_SINGLE_NUMBER_III
{
class Solution
{
public:
vector<int> singleNumber(vector<int>& nums)
{
int xorOfAll = 0;
for (unsigned int i = 0; i < nums.size(); i++)
{
xorOfAll = xorOfAll ^ nums[i];
}
while ((mask & xorOfAll) == 0)
{
}

for (unsigned int i = 0; i < nums.size(); i++)
{
if ((nums[i] & mask) != 0)
{
}
}

return vector<int>(result, result + _countof(result));
}
};
};

using namespace _LEET_SINGLE_NUMBER_III;

int LEET_SINGLE_NUMBER_III()
{
int case1[] = { 1, 2, 3, 2, 1, 5 };
Solution s;
vector<int> result = s.singleNumber(vector<int>(case1, case1+_countof(case1)));
cout << result[0] << ", " << result[1] << endl;
return 0;
}