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]
Example 2:
Input: nums = [1,1] Output: [2]
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.
Companies:
Facebook
Related Topics:
Array, Hash Table
Similar Questions:
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// OJ: https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> findDisappearedNumbers(vector<int>& A) {
int N = A.size();
for (int i = 0; i < N; ++i) {
while (A[A[i] - 1] != A[i]) swap(A[A[i] - 1], A[i]);
}
vector<int> ans;
for (int i = 0; i < N; ++i) {
if (A[i] != i + 1) ans.push_back(i + 1);
}
return ans;
}
};