1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
|
// 260. Single Number III
#include "leetcode.h"
/*
* given an integer array 'nums' in which exactly two elements appear only once
* and all the otherelements appear exactly twice. find the two elements that
* appear only once. you can return the answer in any order. you must write an
* algorithm that runs linear runtime complexity and uses only constant extra
* space.
*/
int *singleNumber(int *nums, int numsSize, int *returnSize) {
*returnSize = 2;
int *ans = (int *)calloc(2, sizeof(int));
long long tmp = 0;
for (int i = 0; i < numsSize; i++)
tmp ^= nums[i];
tmp &= -tmp;
for (int i = 0; i < numsSize; i++)
if (!(tmp & nums[i]))
ans[0] ^= nums[i];
else
ans[1] ^= nums[i];
return ans;
}
int main() {
int n1[] = {1, 2, 1, 3, 2, 5}, n2[] = {-1, 0}, n3[] = {0, 1}, rs1, rs2, rs3;
int *sn1 = singleNumber(n1, ARRAY_SIZE(n1), &rs1); // expect: 3 5
int *sn2 = singleNumber(n2, ARRAY_SIZE(n2), &rs2); // expect: -1 0
int *sn3 = singleNumber(n3, ARRAY_SIZE(n3), &rs3); // expect: 1 0
for (int i = 0; i < rs1; i++)
printf("%d ", sn1[i]);
printf("\n");
for (int i = 0; i < rs2; i++)
printf("%d ", sn2[i]);
printf("\n");
for (int i = 0; i < rs3; i++)
printf("%d ", sn3[i]);
printf("\n");
free(sn1), free(sn2), free(sn3);
}
|