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Java Solution for LeetCode algorithm problems, continually updating
1920\. Build Array from Permutation
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**).
**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)?