[leetcode] 80. Remove Duplicates from Sorted Array II

Remove Duplicates from Sorted Array II

Given an integer array nums sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. The relative order of the elements should be kept the same.

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.

 

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

Example 2:

Input: nums = [0,0,1,1,1,1,2,3,3]
Output: 7, nums = [0,0,1,1,2,3,3,_,_]
Explanation: Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

 

Constraints:

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

 

풀이 - 1

일단 생각나는대로 풀어보았고, 이전문제에서 배열의 길이가 조절되는 언어를 사용하면서 풀이과정에 포함시키지 않아 이번에는 포함시켜 보았다.

function removeDuplicates(nums: number[]): number {
  let count: number = 1;
  let MAX_EXIST_COUNT: number = 2;
  let popCount: number = 0;
  let prev: number = nums[0];
  let cur: number;
  let idx: number = 1;

  for (let i = 1; i < nums.length; i++) {
    cur = nums[i];
    if (prev !== cur) {
      nums[idx] = cur;
      idx++;
      prev = cur;
      count = 1;
    } else if (prev === cur && MAX_EXIST_COUNT > count) {
      nums[idx] = cur;
      idx++;
      count++;
    } else if (prev === cur && MAX_EXIST_COUNT === count) {
      popCount++;
    }
  }

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

  return nums.length;
}

변수와 if문이 좀 지저분해 보여 코드를 정리할 필요가 있어 보인다.

 

풀이 - 2

크게 달라보이진 않는데, 현 수준에서는 이게 최선인것 같다..

function removeDuplicates(nums: number[]): number {
  let MAX_EXIST_COUNT: number = 2;

  let count: number = 1;
  let popCount: number = 0;

  let prev: number = nums[0];
  let cur: number;
  
  let idx: number = 1;

  for (let i = 1; i < nums.length; i++) {
    cur = nums[i];

    if (prev !== cur) {
      nums[idx] = cur;
      idx++;
      prev = cur;
      count = 1;
    } else if (prev === cur) {
      if (MAX_EXIST_COUNT > count) {
        nums[idx] = cur;
        idx++;
        count++;
      } else {
        popCount++;
      }
    }
  }

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

  return nums.length;
}

같은 조건안에서 달라지는 조건을 서로 적용하고 변수들을 보기쉽게 나눠놓았는데 성능에는 큰 차이 없는것 같다..