1920. 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]```

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)?

1920. Build Array from Permutation
``````struct Solution;

impl Solution {
fn build_array(nums: Vec<i32>) -> Vec<i32> {
let n = nums.len();
let mut res = vec![0; n];
for i in 0..n {
res[i] = nums[nums[i] as usize];
}
res
}
}
#[test]
fn test() {
let nums = vec![0, 2, 1, 5, 3, 4];
let res = vec![0, 1, 2, 4, 5, 3];
assert_eq!(Solution::build_array(nums), res);
let nums = vec![5, 0, 1, 2, 3, 4];
let res = vec![4, 5, 0, 1, 2, 3];
assert_eq!(Solution::build_array(nums), res);
}
``````