# 1929. Concatenation of Array

## Description

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

## Solutions

• class Solution {
public int[] getConcatenation(int[] nums) {
int n = nums.length;
int[] ans = new int[n << 1];
for (int i = 0; i < n << 1; ++i) {
ans[i] = nums[i % n];
}
return ans;
}
}

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

• class Solution:
def getConcatenation(self, nums: List[int]) -> List[int]:
return nums + nums


• func getConcatenation(nums []int) []int {
return append(nums, nums...)
}

• function getConcatenation(nums: number[]): number[] {
return [...nums, ...nums];
}


• /**
* @param {number[]} nums
* @return {number[]}
*/
var getConcatenation = function (nums) {
let ans = nums.slice();
ans.splice(nums.length, 0, ...nums);
return ans;
};


• impl Solution {
pub fn get_concatenation(nums: Vec<i32>) -> Vec<i32> {
nums.repeat(2)
}
}