Table of contents
Given problem
Given an array nums of n
integers where nums[i]
is in the range [1, n]
, return an array of all the integers in the range [1, n]
that do not appear in nums.

Example 1:
 Input: nums =
[4,3,2,7,8,2,3,1]
 Output:
[5,6]
 Input: nums =

Example 2:
 Input: nums =
[1,1]
 Output:
[2]
 Input: nums =

Constraints:
n == nums.length
1 <= n <= 105
1 <= nums[i] <= n
Follow up: Could you do it without extra space and in O(n)
runtime? You may assume the returned list does not count as extra space.
Using Cyclic Sort
To know more other ways to solve this issue, we can refer to the following article Leetcode 268  Missing Number.
Currently, we will apply the Cyclic Sort. Then, iterate each element in an array to check the difference between its value and its index.
Below is the source code of this solution:
class Solution {
public List<Integer> findDisappearedNumbers(int[] nums) {
int size = nums.length;
int start = 0;
while (start < size) {
int current = nums[start]  1;
if (nums[start] != nums[current]) {
swap(nums, start, current);
} else {
++start;
}
}
List<Integer> missingNumbers = new ArrayList<>();
for (int i = 0; i < size; ++i) {
if (nums[i] != i + 1) {
missingNumbers.add(i + 1);
}
}
return missingNumbers;
}
private static void swap(int[] nums, int i, int j) {
int tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
}
}
The complexity of this solution is:
 Time complexity: O(n)
 Space complexity: O(1)
Wrapping up
Refer: