# LeetCode 1920
###### tags: `python` `LeetCode`
>LeetCode 第1920解題
>使用Python

## 題目

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]

## 解題思路

 nums = [0,2,1,5,3,4]
 ans  = [0,1,2,4,5,3]
 n is the lenth of nums
 ans[0] = nums[0]
 .
 .
 .
 ans[n-1] = nums[n-1]
 
#1. idx = 0, ans[0] = nums[0] = 0
#2. idx = 2, ans[1] = nums[2] = 2
#6. idx = 4, ans[5] = nums[4] = 3

## 程式碼

def buildArray(nums: List[int]):
    ans = []
    for idx in nums:
        ans.append(nums[idx])
    return ans

###### 作者:@allianRZ