Question

1929. Concatenation of Array

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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

Solution 1

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

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;
  }
};

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

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

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

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

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* getConcatenation(int* nums, int numsSize, int* returnSize) {
  int* ans = malloc(sizeof(int) * numsSize * 2);
  for (int i = 0; i < numsSize; i++) {
    ans[i] = ans[i + numsSize] = nums[i];
  }
  *returnSize = numsSize * 2;
  return ans;
}