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// 442. Find All Duplicates in an Array
#include "leetcode.h"
/*
* 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 appear twice. you must write an algorithm that
* runs in 'O(n)' time and uses only constant extra space.
*/
int *findDuplicates(int *nums, int numsSize, int *returnSize) {
*returnSize = 0;
if (!nums || numsSize < 2)
return 0;
int *ans = NULL;
for (int i = 0; i < numsSize; i++) {
int n = abs(nums[i]) - 1;
if (nums[n] < 0) {
(*returnSize)++;
ans = realloc(ans, sizeof(int) * *returnSize);
ans[*returnSize - 1] = abs(nums[i]);
} else
nums[n] = -1 * nums[n];
}
return ans;
}
int main() {
int n1[] = {4, 3, 2, 7, 8, 2, 3, 1}, n2[] = {1, 1, 2}, n3[] = {1}, rs1, rs2,
rs3;
int *fd1 = findDuplicates(n1, ARRAY_SIZE(n1), &rs1);
int *fd2 = findDuplicates(n2, ARRAY_SIZE(n2), &rs2);
int *fd3 = findDuplicates(n3, ARRAY_SIZE(n3), &rs3);
if (fd1)
for (int i = 0; i < rs1; i++)
printf("%d ", fd1[i]);
else
printf("NULL");
printf("\n");
if (fd2)
for (int i = 0; i < rs2; i++)
printf("%d ", fd2[i]);
else
printf("NULL");
printf("\n");
if (fd3)
for (int i = 0; i < rs3; i++)
printf("%d ", fd3[i]);
else
printf("NULL");
printf("\n");
free(fd1), free(fd2), free(fd3);
}
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