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
43
44
45
46
47
48
|
// 442. Find All Duplicates in an Array
#include <math.h>
#include <stdio.h>
#include <stdlib.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 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");
}
|