[leetcode] 26. Remove Duplicates from Sorted Array

Remove Duplicates from Sorted Array

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums.

Consider the number of unique elements of nums to 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 unique elements in the order they were present in nums initially. 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[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
    assert nums[i] == expectedNums[i];
}

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

 

Example 1:

Input: nums = [1,1,2]
Output: 2, nums = [1,2,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

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

 

Constraints:

• 1 <= nums.length <= 3 * 104
• -100 <= nums[i] <= 100
• nums is sorted in non-decreasing order.

 

풀이 - 1

배열안에 있는 요소들의 중복을 제거하는 문제로 간단하게 Set 자료구조의 특성을 적용해보았다.

function removeDuplicates(nums: number[]): number {
  const arr = [...new Set(nums)];

  for (let i = 0; i < nums.length; i++) {
    if (i < arr.length) {
      nums[i] = arr[i];
    }
  }

  return arr.length;
}

아무래도 정답은 정답이지만 new Set으로 자료구조를 변형하고 또한번 배열로 하다보니 시간복잡도상 좋지 않아 보인다.

 

풀이 - 2

배열을 한번 순회하면서 정답이 나오도록 최대한 생각해봤다.

function removeDuplicates(nums: number[]): number {
  let prev: number = nums[0];
  let count: number = 1;

  for (let i = 1; i < nums.length; i++) {
    if (prev !== nums[i]) {
      nums[count] = nums[i];
      count++;
      prev = nums[i];
    }
  }

  return count;
}

리턴 값으로는 count라고 변수명을 짓는게 맞는거 같다가도 할당할때 index 대신 사용하는게 과연 맞는걸까 라는 의문점이 남지만 일단 구현하는데에는 성공 !