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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.

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Related Topics:
Array, Hash Table

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Solution 1.

// 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;
    }
};