#+title: 136 Single Number

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

You must implement a solution with a linear runtime complexity and use only constant extra space.



Example 1:

Input: nums = [2,2,1]

Output: 1

Example 2:

Input: nums = [4,1,2,1,2]

Output: 4

Example 3:

Input: nums = [1]

Output: 1



Constraints:

    1 <= nums.length <= 3 * 104
    -3 * 104 <= nums[i] <= 3 * 104
    Each element in the array appears twice except for one element which appears only once.

* Solution

#+name: solution
#+begin_src C
int xor(int a, int b) {
   return a ^ b;
}

int singleNumber(int* nums, int numsSize) {
    int number = 0;
    for (int i=0; i < numsSize; i++) {
        number = xor(number, nums[i]);
    }
    return number;
}
#+end_src

#+RESULTS: solution

#+begin_src C :noweb yes
#include <stdio.h>

<<solution>>

int numbers[3] = {1, 1, 2};
int single = singleNumber(numbers, 3);
printf("%d\n", single);
#+end_src

#+RESULTS:
: -2