27. Remove Element

# Description

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the elements may be changed. Then return the number of elements in nums which are not equal to val.

Consider the number of elements in nums which are not equal to val be k, to get accepted, you need to do the following things:

Change the array nums such that the first k elements of nums contain the elements which are not equal to val. The remaining elements of nums are not important as well as the size of nums.
Return k.

Custom Judge:

The judge will test your solution with the following code:

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int[] nums = [...]; // Input array
int val = ...; // Value to remove
int[] expectedNums = [...]; // The expected answer with correct length.
                            // It is sorted with no values equaling val.

int k = removeElement(nums, val); // Calls your implementation

assert k == expectedNums.length;
sort(nums, 0, k); // Sort the first k elements of nums
for (int i = 0; i < actualLength; i++) {
    assert nums[i] == expectedNums[i];
}

If all assertions pass, then your solution will be accepted.

Example 1:

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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).

Example 2:

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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

# 풀이

val값을 제외하기 위해 nums를 순회하며 val값이 아니라면, 0번째 인덱스(idx)부터 nums[i]를 넣어 nums를 수정했다.
만약, nums = [3, 2, 2, 3] 이고 val = 3이라면
nums[0] == 3이고 nums[1] != 3 이기 때문에 nums[0]idx = 0 에는 nums[1]이 들어가게 된다.
이후 nums[2] != val 이니 nums[1]idx = 1에 nums[2]가 들어간다.

# 코드

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class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        idx = 0
        for i in range(len(nums)):
            if nums[i] != val:
                nums[idx] = nums[i]
                idx += 1
        return idx