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.length1 <= n <= 10001 <= nums[i] <= 1000
class Solution:
def getConcatenation(self, nums: List[int]) -> List[int]:
return nums + numsclass 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;
}
}/**
* @param {number[]} nums
* @return {number[]}
*/
var getConcatenation = function (nums) {
let ans = nums.slice();
ans.splice(nums.length, 0, ...nums);
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;
}
};func getConcatenation(nums []int) []int {
return append(nums, nums...)
}