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);
}