[leetcode] 27. Remove Element

Remove Element

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:

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:

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:

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 - 1

머리가 말랑말랑 해졌나.. 한번에 맞췄다.

function removeElement(nums: number[], val: number): number {
  let idx = 0;
  let popCount = 0;

  for (let i = 0; i < nums.length; i++) {
    if (nums[i] !== val) {
      nums[idx] = nums[i];
      idx++;
    } else {
      popCount++;
    }
  }

  for (let y = 0; y < popCount; y++) {
    nums.pop();
  }

  return nums.length;
}

실제 할당할 idx와 필요없는 배열을 지울 popCount를 선언하여 입력받은 val이 해당하지 않을때만 idx를 증가시키고 해당 요소에 할당하는 식으로 풀이하였다.