1920. Build Array from Permutation

Method 1

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]

Method 1

【O(n) time | O(n) space】

package Leetcode.Array;

/**
 * @author zhengxingxing
 * @date 2025/05/06
 */
public class BuildArrayFromPermutation {
    
    public static int[] buildArray(int[] nums) {
        int n = nums.length;
        // Create result array with same length as input array
        int[] ans = new int[n];

        // Iterate through array and build result
        // For each position i, ans[i] = nums[nums[i]]
        for(int i = 0; i < n; i++) {
            ans[i] = nums[nums[i]];
        }

        return ans;
    }

    public static void main(String[] args) {
        // Test Case 1: Expected output [0,1,2,4,5,3]
        int[] nums1 = {0,2,1,5,3,4};
        int[] result1 = buildArray(nums1);
        System.out.print("Test Case 1 Result: ");
        for(int num : result1) {
            System.out.print(num + " ");
        }
        System.out.println();

        // Test Case 2: Expected output [4,5,0,1,2,3]
        int[] nums2 = {5,0,1,2,3,4};
        int[] result2 = buildArray(nums2);
        System.out.print("Test Case 2 Result: ");
        for(int num : result2) {
            System.out.print(num + " ");
        }
    }
}

Method 1

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