[LeetCode] Remove Element

Question

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The relative order of the elements may be changed.
Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums.
More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result.
It does not matter what you leave beyond the first k elements.
Return k after placing the final result in the first k slots of nums.
Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

하나의 리스트(nums1)와 정수값 val이 주어진다.
리턴 값 없이, nums1에서 val과 일치하는 값을 제거하는 문제였다.

Example

Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2,_,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 2.
It does not matter what you leave beyond the returned k (hence they are underscores).

Input: nums = [0,1,2,2,3,0,4,2], val = 2
Output: 5, nums = [0,1,4,0,3,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.
Note that the five elements can be returned in any order.
It does not matter what you leave beyond the returned k (hence they are underscores).

Constraints

• 0 <= nums.length <= 100
• 0 <= nums[i] <= 50
• 0 <= val <= 100

Solution

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        res=[]
        for x in nums:
            if x!=val:
                res.append(x)
                
        nums.clear()
        for x in res:
            nums.append(x)
        

• 직접 푼 풀이이다.
• res 리스트를 하나 더 만들어서 nums에서 val과 일치하지 않는 값만 res에 집어넣는다.
• nums를 초기화 한 뒤에 res에 있는 값을 집어넣는다.

끄적

• 영어로 적혀 있어서 문제를 해석하는 것 보다 빠르게 예제를 확인하는게 나았다.

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