136. Single Number

Given an 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?

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<int>::size_type i;
        for(i = 0; i < nums.size() - 2; i += 2)
        {
            if(nums[i] != nums[i + 1])
            {
                break;
            }
        }
        return nums[i];
    }
};

排序可以过，但是不满足时间复杂度O(n)的条件
讨论区一个很巧妙的算法，利用亦或

https://discuss.leetcode.com/topic/1916/my-o-n-solution-using-xor

^ 按位异或 若参加运算的两个二进制位值相同则为0，否则为1

对于如下：

int x = 5;

int y = 0;

cout << (x ^ y);//5

y = 5;

cout << (x ^ y);//0

题外话：利用亦或，不用第三个变量交换两个数。

int main()
{
	int x = 10;
    int y = 100;
    cout << "x = " << x << ";y = " << y << endl;

    x = x ^ y;
    y = x ^ y;
    x = x ^ y;

    cout << "x = " << x << ";y = " << y << endl;
	return 0;
}

即：

x = x ^ y;

y = x ^ y = x ^ y ^ y = x;

x = x ^ y = x ^ y ^ x = y;

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;

class Solution {
public:
    int singleNumber(vector<int>& nums) {
    	int result = 0;
    	for (auto iter = nums.begin(); iter != nums.end(); iter++)
    	{
    		result ^= *iter;
    	}
    	return result;
    }
};

int main()
{
	Solution s;
	vector<int>vec{1,2,3,3,5,2,1,4,4};
	cout << s.singleNumber(vec);
	return 0;
}

相同的题：

389. Find the Difference

Given two strings s and t which consist of only lowercase letters.

String t is generated by random shuffling string s and then add one more letter at a random position.

Find the letter that was added in t.

Example:

Input:
s = "abcd"
t = "abcde"

Output:
e

Explanation:
'e' is the letter that was added.

class Solution {
public:
    char findTheDifference(string s, string t) {
    	char ch = 0;
        for (string::size_type i = 0; i < s.size(); ++i)
        {
        	ch ^= s[i];
        }
        for (string::size_type i = 0; i < t.size(); ++i)
        {
        	ch ^= t[i];
        }
        return ch;
    }
};