README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1900-1999/1929.Concatenation%20of%20Array/README_EN.md	1132	Weekly Contest 249 Q1	Array Simulation

1929. Concatenation of Array

中文文档

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

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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

TypeScript

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

Rust

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

JavaScript

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

C

/**
 * 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;
}