BuildArrayFromPermutation1920.kt

package 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).

 

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.
**/
/*
Leetcode link:
https://leetcode.com/problems/build-array-from-permutation/
*/

   fun buildArray(nums: IntArray): IntArray {
        val numbers =IntArray(nums.size){0}
        nums.forEachIndexed{i,num->
          numbers[i]= nums[num]
        }
        return numbers
    }