array-from-permutation.cpp

//question//

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)?


//solution//

class Solution {
public:
    vector<int> buildArray(vector<int>& nums) {
        vector<int> v;
        for(int i=0;i<nums.size();i++)
        {
            v.push_back(nums[nums[i]]);
        }
        return v;
    }
};