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1920. Build Array from Permutation

Description

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)?

Solutions

Solution 1

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class Solution:
    def buildArray(self, nums: List[int]) -> List[int]:
        return [nums[num] for num in nums]

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class Solution {
    public int[] buildArray(int[] nums) {
        int[] ans = new int[nums.length];
        for (int i = 0; i < nums.length; ++i) {
            ans[i] = nums[nums[i]];
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> buildArray(vector<int>& nums) {
        vector<int> ans;
        for (int& num : nums) {
            ans.push_back(nums[num]);
        }
        return ans;
    }
};

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func buildArray(nums []int) []int {
    ans := make([]int, len(nums))
    for i, num := range nums {
        ans[i] = nums[num]
    }
    return ans
}

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function buildArray(nums: number[]): number[] {
    return nums.map(v => nums[v]);
}

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impl Solution {
    pub fn build_array(nums: Vec<i32>) -> Vec<i32> {
        nums.iter().map(|&v| nums[v as usize]).collect()
    }
}

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/**
 * @param {number[]} nums
 * @return {number[]}
 */
var buildArray = function (nums) {
    let ans = [];
    for (let i = 0; i < nums.length; ++i) {
        ans[i] = nums[nums[i]];
    }
    return ans;
};

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/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* buildArray(int* nums, int numsSize, int* returnSize) {
    int* ans = malloc(sizeof(int) * numsSize);
    for (int i = 0; i < numsSize; i++) {
        ans[i] = nums[nums[i]];
    }
    *returnSize = numsSize;
    return ans;
}

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