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Java-based LeetCode algorithm problem solutions, regularly updated

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1920\. Build Array from Permutation

Easy

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]

**Constraints:**

*   `1 <= nums.length <= 1000`
*   `0 <= nums[i] < nums.length`
*   The elements in `nums` are **distinct**.

**Follow-up:** Can you solve it without using an extra space (i.e., `O(1)` memory)?