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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

Solution 1: Simulation

We directly simulate according to the problem description by adding the elements of $\textit{nums}$ to the answer array one by one, and then adding the elements of $\textit{nums}$ to the answer array again.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.

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

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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;
    }
}

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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;
    }
};

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

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

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

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/**
 * @param {number[]} nums
 * @return {number[]}
 */
var getConcatenation = function (nums) {
    return [...nums, ...nums];
};

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/**
 * 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;
}

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