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author | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
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committer | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
commit | d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch) | |
tree | 96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /single_number3.c | |
download | leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.tar.gz leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.zip |
initial
Diffstat (limited to 'single_number3.c')
-rw-r--r-- | single_number3.c | 42 |
1 files changed, 42 insertions, 0 deletions
diff --git a/single_number3.c b/single_number3.c new file mode 100644 index 0000000..16a24bb --- /dev/null +++ b/single_number3.c @@ -0,0 +1,42 @@ +// 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); +} |