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?

題目

給定一個整數數組，里面只有一個數只出現一次，其余的數都出現兩次，找出只出現一次的數。

方法

采用異或運算^
a ^ a = 0
a ^ 0 = a
a ^ b ^ c = a ^ (b ^ c)

c代碼

#include <assert.h>

int singleNumber(int* nums, int numsSize) {
    int i = 0;
    int single = 0;
    for(i = 0; i < numsSize; i++) {
        single ^= nums[i];
    }
    return single;
}

int main() {
    int nums[5] = {1,3,3,1,5};
    assert(singleNumber(nums, 5) == 5);

    return 0;
}