LeetCode : Single Number (C++)

Given a non-empty array of integers, every element appears twice except for one. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

Input: [2,2,1]
Output: 1

Example 2:

Input: [4,1,2,1,2]
Output: 4

Solution :

	#include <iostream>
	#include <vector>
	#include <algorithm>
	#include <set>
	class Solution
	{
	public:
	static int singleNumber( std::vector<int> nums )
	{
	unsigned int result = 0;
	// Get all unique number and add them
	std::set<int> mSet( nums.begin(), nums.end() );
	for( std::set<int>::iterator it = mSet.begin(); it != mSet.end(); it++ )
	{
	result += *it;
	}
	// we must have pairs , so double it
	result = 2 * result;
	// Reduce all the number to find what's left that must be the
	// un-matched number
	for( unsigned int i = 0; i < nums.size(); i++ )
	{
	result -= nums[i];
	}
	return result;
	}
	};
	int main()
	{
	std::vector<int> nums = { 4, 1, 2, 1, 2 };
	// should return 4 for this case
	std::cout << Solution::singleNumber( nums ) << std::endl;
	return 0;
	}

view raw SingleNumber.cpp hosted with ❤ by GitHub