442. Find All Duplicates in an Array

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]

idea: similar methods with 448 Find All Numbers Disappeared in an Array

Solution 1: If nums[index] is already <0, index+1 is the duplicated number

class Solution {
public:
    vector<int> findDuplicates(vector<int>& nums) {
        int n=nums.size();
        vector<int> res;
        for (int i=0;i<n;i++){
            int index=abs(nums[i])-1; 
            if (nums[index]<0) res.push_back(abs(nums[i])); //diff with 448
            else nums[index]=-nums[index];
        }
        return res;
    }
};

Solution 2: If nums[i]!= i+1, nums[i] is the duplicated number.

class Solution {
public:
    vector<int> findDuplicates(vector<int>& nums) {
        int n=nums.size();
        vector<int> res;
        for (int i=0;i<n;i++){
            int index=nums[i]-1;
            if (nums[index]!=nums[i]) {
                swap (nums[i],nums[index]);
                --i;
            }
        }
        for (int i=0;i<n;i++){
            if (nums[i]!=i+1) res.push_back(nums[i]); //diff with 448
        }
        return res;
    }
};

Solution 3: If nums[i]>2*n, i+1 is the duplicated number.

class Solution {
public:
    vector<int> findDuplicates(vector<int>& nums) {
        int n=nums.size();
        vector<int> res;
        for (int i=0;i<n;i++){
            nums[(nums[i]-1)%n]+=n;
        }
        for (int i=0;i<n;i++){
            if (nums[i]>2*n) res.push_back(i+1); //diff with 448
        }
        return res;
    }
};