1.描述
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?
2.分析
x ^ x == 0
3.代码
int singleNumber(int* nums, int numsSize) {
if (NULL == nums || numsSize <= 0) exit(-1);
int single = nums[0];
for (unsigned int i = 1; i < numsSize; ++i) {
single ^= nums[i];
}
return single;
}
