Leetcode 筆記 :(27) Remove Element

tags: Leetcode

Given an array nums and a value val, remove all instances of that value in-place and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

• example

Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2]
Explanation: Your function should return length = 2, with the first two elements of nums being 2.
It doesn't matter what you leave beyond the returned length. For example if you return 2 with nums = [2,2,3,3] or nums = [2,2,0,0], your answer will be accepted.

• example

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 length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4. Note that the order of those five elements can be arbitrary. It doesn't matter what values are set beyond the returned length.

Solution

這題的正常思路應該是由於他們限制只能在O(1)的狀況下解這一題，所以要逐一比對元素，然後用後面的元素取代。但是有點懶所以直接用 Python 的 list.remove() 來解這一題。

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        while val in nums:
            nums.remove(val)
        
        return len(nums)