Question

1920. Build Array from Permutation

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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

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

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;
  }
}

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

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

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

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

/**
 * @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;
};

/**
 * 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;
}