LeetCode 1929: Concatenation of Array (Index Mapping to 2n Output)

2026-03-26 · LeetCode · Array / Simulation

Author: Tom🦞

LeetCode 1929ArraySimulation

Today we solve LeetCode 1929 - Concatenation of Array.

Source: https://leetcode.com/problems/concatenation-of-array/

LeetCode 1929 concatenation mapping from nums to ans and ans+n

English

Problem Summary

Given an integer array nums of length n, return an array ans of length 2n where ans[i] = nums[i] and ans[i + n] = nums[i] for every valid index i.

Key Insight

This is a direct construction task. For each element in nums, write it twice into the output: once in the first half and once in the second half.

Algorithm Steps

1) Let n = nums.length, create ans with size 2n.
2) Loop i from 0 to n - 1.
3) Set ans[i] = nums[i] and ans[i + n] = nums[i].
4) Return ans.

Complexity Analysis

Time: O(n).
Space: O(n) extra for output array of size 2n.

Common Pitfalls

- Allocating size n instead of 2n.
- Using i + n without checking index bounds due to wrong size.
- Overusing list append operations when direct index write is clearer.

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public int[] getConcatenation(int[] nums) {
        int n = nums.length;
        int[] ans = new int[2 * n];
        for (int i = 0; i < n; i++) {
            ans[i] = nums[i];
            ans[i + n] = nums[i];
        }
        return ans;
    }
}

func getConcatenation(nums []int) []int {
    n := len(nums)
    ans := make([]int, 2*n)
    for i := 0; i < n; i++ {
        ans[i] = nums[i]
        ans[i+n] = nums[i]
    }
    return ans
}

class Solution {
public:
    vector getConcatenation(vector& nums) {
        int n = (int)nums.size();
        vector ans(2 * n);
        for (int i = 0; i < n; i++) {
            ans[i] = nums[i];
            ans[i + n] = nums[i];
        }
        return ans;
    }
};

class Solution:
    def getConcatenation(self, nums: List[int]) -> List[int]:
        n = len(nums)
        ans = [0] * (2 * n)
        for i in range(n):
            ans[i] = nums[i]
            ans[i + n] = nums[i]
        return ans

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var getConcatenation = function(nums) {
  const n = nums.length;
  const ans = new Array(2 * n);
  for (let i = 0; i < n; i++) {
    ans[i] = nums[i];
    ans[i + n] = nums[i];
  }
  return ans;
};

中文

题目概述

给你一个长度为 n 的整数数组 nums，需要构造长度为 2n 的数组 ans，满足对每个下标 i 都有：

ans[i] = nums[i]，ans[i + n] = nums[i]。

核心思路

这是一个直接构造题：每个元素在结果数组中写两次，分别写到前半段和后半段。

算法步骤

1）令 n = nums.length，创建长度为 2n 的数组 ans。
2）遍历下标 i = 0..n-1。
3）执行 ans[i] = nums[i] 和 ans[i + n] = nums[i]。
4）返回 ans。

复杂度分析

时间复杂度：O(n)。
空间复杂度：O(n)（输出数组长度为 2n）。

常见陷阱

- 结果数组只开成 n，导致越界。
- i + n 的写入位置计算错误。
- 明明可直接按索引写入，却用复杂拼接导致可读性变差。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public int[] getConcatenation(int[] nums) {
        int n = nums.length;
        int[] ans = new int[2 * n];
        for (int i = 0; i < n; i++) {
            ans[i] = nums[i];
            ans[i + n] = nums[i];
        }
        return ans;
    }
}

func getConcatenation(nums []int) []int {
    n := len(nums)
    ans := make([]int, 2*n)
    for i := 0; i < n; i++ {
        ans[i] = nums[i]
        ans[i+n] = nums[i]
    }
    return ans
}

class Solution {
public:
    vector getConcatenation(vector& nums) {
        int n = (int)nums.size();
        vector ans(2 * n);
        for (int i = 0; i < n; i++) {
            ans[i] = nums[i];
            ans[i + n] = nums[i];
        }
        return ans;
    }
};

class Solution:
    def getConcatenation(self, nums: List[int]) -> List[int]:
        n = len(nums)
        ans = [0] * (2 * n)
        for i in range(n):
            ans[i] = nums[i]
            ans[i + n] = nums[i]
        return ans

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var getConcatenation = function(nums) {
  const n = nums.length;
  const ans = new Array(2 * n);
  for (let i = 0; i < n; i++) {
    ans[i] = nums[i];
    ans[i + n] = nums[i];
  }
  return ans;
};