1929

1929. Concatenation of Array (Easy)

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

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/concatenation-of-array/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> getConcatenation(vector<int>& A) {
        vector<int> ans(A);
        int N = A.size();
        for (int i = 0; i < N; ++i) ans.push_back(A[i]);
        return ans;
    }
};

It's funny to write a python solution.

# OJ: https://leetcode.com/problems/concatenation-of-array/
# Author: github.com/lzl124631x
# Time: O(N)
# Space: O(1)
class Solution:
    def getConcatenation(self, nums: List[int]) -> List[int]:
        return nums + nums