// 442. Find All Duplicates in an Array #include #include #include /* * given an integer array 'nums' of length 'n' where all the integers of 'nums' * are in the range [1, n] and each integer appears once or twice. return an * array of all the integers that appears twice. you must write an algorithm * that run O(n) time and uses only constant extra space. */ int *findDuplicats(int *nums, int nums_size, int *return_size) { if (!nums || nums_size < 2) { *return_size = 0; return 0; } int *ans = NULL; *return_size = 0; for (int i = 0; i < nums_size; i++) { int ti = abs(nums[i]) - 1; if (nums[ti] < 0) { *return_size += 1; ans = realloc(ans, sizeof(int)); ans[*return_size - 1] = abs(nums[i]); } else nums[ti] = -1 * nums[ti]; } return ans; } int main() { int n1[] = {4, 3, 2, 7, 8, 2, 3, 1}, ns1 = 8, rs1[] = {}; int n2[] = {1, 1, 2}, ns2 = 3, rs2[] = {}; int n3[] = {1}, ns3 = 1, rs3[] = {}; int *fd1 = findDuplicats(n1, ns1, rs1); int *fd2 = findDuplicats(n2, ns2, rs2); int *fd3 = findDuplicats(n3, ns3, rs3); for (int i = 0; i < ns1; i++) printf("%d ", fd1[i]); // expect: 2 3 printf("\n"); for (int i = 0; i < ns2; i++) printf("%d ", fd2[i]); // expect: 1 printf("\n"); for (int i = 0; i < ns3; i++) printf("%d ", fd3[i]); // expect: printf("\n"); }