Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements that appear twice in this array.
Could you do it without extra space and in O(n) runtime?
Example:
Input:
[4,3,2,7,8,2,3,1]
Output:
[2,3]

思路：

The concept here is to negate each number‘s index as the input is 1 <= a[i] <= n (n = size of array).

Once a value is negated, if it requires to be negated again then it is a duplicate.

vector<int> findDuplicates(vector<int>& nums)
     {
         vector<int>ret;
         for (int i = 0; i < nums.size();i++)
         {
             int index = abs(nums[i]) - 1;
             
 
if (nums[index] > 0)
             {
                 nums[index] = - nums[index];
             }
             else
             {
                 ret.push_back(abs(nums[i]));
             }
         }
         return ret;
     }

參考：

https://discuss.leetcode.com/topic/64805/java-easy-to-understand-solution-without-extra-space-and-in-o-n-time

[leetcode-442-Find All Duplicates in an Array]