Build Array from Permutation

Given a zero-based permutation nums (0-indexed), build an array ans of the same length where ans[i] = nums[nums[i]] for each 0 <= i < nums.length and return it.

A zero-based permutation nums is an array of distinct integers from 0 to nums.length - 1 (inclusive).

Example 1:

Input: nums = [0,2,1,5,3,4]
Output: [0,1,2,4,5,3]
Explanation: The array ans is built as follows: 
ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]
    = [nums[0], nums[2], nums[1], nums[5], nums[3], nums[4]]
    = [0,1,2,4,5,3]

Example 2:

Input: nums = [5,0,1,2,3,4]
Output: [4,5,0,1,2,3]
Explanation: The array ans is built as follows:
ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]
    = [nums[5], nums[0], nums[1], nums[2], nums[3], nums[4]]
    = [4,5,0,1,2,3]

Approach

Java

import java.util.Arrays;

public class BuildArrayPermut {
    public static void main(String[] args) {

        int[] nums = { 0, 2, 1, 5, 3, 4 };

        int[] ans = buildArray(nums);
        System.out.println(Arrays.toString(ans));

    }

    static int[] buildArray(int[] nums) {
        int ans[] = new int[nums.length];

        for (int i = 0; i < nums.length; i++) {
            ans[i] = nums[nums[i]];
        }
        return ans;
    }

}

C++

#include <bits/stdc++.h>
using namespace std;

vector<int> buildArray(vector<int> &nums)
{
    vector<int> ans(nums.size());

    for (int i = 0; i < nums.size(); i++)
    {
        ans[i] = nums[nums[i]];
    }
    return ans;
}

int main()
{
    vector<int> nums = {0, 2, 1, 5, 3, 4};

    vector<int> ans = buildArray(nums);

    cout << "[";
    for (int i = 0; i < ans.size(); i++)
    {
        cout << ans[i];
        if (i != ans.size() - 1)
            cout << ", ";
    }
    cout << "]";

    return 0;
}