Question
Given an array of numbers nums
, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.
For example:
Given nums = [1, 2, 1, 3, 2, 5]
, return [3, 5]
.
Note:
- The order of the result is not important. So in the above example,
[5, 3]
is also correct. - Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?
Solution
Result: Accepted Time: 8 ms
Here should be some explanations.
/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int* singleNumber(int* nums, int numsSize, int* returnSize) {
int * ans = malloc(sizeof(int) * 2);
*returnSize = 2;
int last_bit = 0, a = 0, b = 0;
for(int i = 0; i < numsSize; i++)
last_bit ^= nums[i];
last_bit = last_bit&(-last_bit);
for(int i = 0; i < numsSize; i++)
if(nums[i] & last_bit)
a ^= nums[i];
else
b ^= nums[i];
ans[0] = a;
ans[1] = b;
return ans;
}
Complexity Analytics
- Time Complexity:
- Space Complexity: