1920. Build Array from Permutation

👩‍💻 문제 (link)

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]

 

 

 

💡 풀이

class Solution:
    def buildArray(self, nums: List[int]) -> List[int]:
        ans =[]
        for i in range(len(nums)):
            ans.append(nums[nums[i]])
        return ans

 

 

✍️ 해설

쉬움. 그냥 빈 리스트 만들어서 조건에 맞게 append 새로 해주면 됨

 

🥳 배운점

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