(Leet Code c++)Concatenation of Array

1929. Concatenation of Array

 

Given an integer array nums of length n, you want to create an array ans of length 2n where ans[i] == nums[i] and ans[i + n] == nums[i] for 0 <= i < n (0-indexed).

Specifically, ans is the concatenation of two nums arrays.

Return the array ans.

 

Example 1:

Input: nums = [1,2,1] Output: [1,2,1,1,2,1] Explanation: The array ans is formed as follows: - ans = [nums[0],nums[1],nums[2],nums[0],nums[1],nums[2]] - ans = [1,2,1,1,2,1]

Example 2:

Input: nums = [1,3,2,1] Output: [1,3,2,1,1,3,2,1] Explanation: The array ans is formed as follows: - ans = [nums[0],nums[1],nums[2],nums[3],nums[0],nums[1],nums[2],nums[3]] - ans = [1,3,2,1,1,3,2,1]

 

Constraints:

• n == nums.length
• 1 <= n <= 1000
• 1 <= nums[i] <= 1000

 

class Solution {
public:
    vector<int> getConcatenation(vector<int>& nums) {
        int n = nums.size();
        vector<int> answer(n * 2, 0);
        
        for(int i = 0; i < nums.size(); i++){
            answer[i] = nums[i];
            answer[i + n] = nums[i];
        }
        
        return answer;
    }
};