Find all missing numbers in the array

We can mark the array directly , If there has been , We can turn it into a negative number , If it's negative , Then it means that this number often appears , Greater than 0 It means that there has been no .

step ：

1. Traverse each number in the entire array , If not, multiply by -1, Here, the array subscript represents each number in the array

Be careful ： because vector The subscript in is from 0 To n-1, The data range is from 1 To n So counting to the array requires -1 go in , Then there may be points appearing twice , Therefore, the number of current traversal should be changed into a positive number in advance ;

2. Traversal array , If there is an array, the following table is positive , It means there is no such thing as , We will i+1, Save to array .

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
       for(auto c:nums){
           c=abs(c);
           if(nums[c-1]>0)nums[c-1]*=-1;
       }
       vector<int>res;
       for(int i=0;i<nums.size();i++){
           if(nums[i]>0)res.push_back(i+1);

       }
       return res;
    }
};