Java, LeetCode | 26. Remove Duplicates from Sorted Array

🚀 풀이 후기

정답인지 검사할 때, 함수의 반환 자료형인 int 의 값과, 인풋이었던 nums 를 모두 확인한다는 부분이 신기했습니다.

🌒 문제 설명

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.

🌓 문제 풀이

class Solution {
    public int removeDuplicates(int[] nums) {
    	// 중복을 제거하기 위해 HashSet 자료형을 사용합니다.
        Set<Integer> set = new HashSet<>();
        
        // for문을 돌면서 nums 배열의 요소를 하나씩 set에 저장합니다.
        for(int num: nums){
            set.add(num);
        }
        
        // 중복 숫자가 제거된 set으로 ArrayList를 만듭니다.
        List<Integer> answerList = new ArrayList<>(set);
        
        // ArrayList를 정렬합니다.
        Collections.sort(answerList);
    
        // ArrayList의 내용을 다시 nums 배열에 넣습니다.
        for(int i=0;i<answerList.size();i++){
            nums[i]=answerList.get(i);
        }
        
        // 중복이 제거된 set의 사이즈를 반환합니다.
        return set.size();
    }
}

🌕 리팩토링

set을 사용하지 않고, nums배열을 순환하면서 바로 중복 숫자가 제거된 수를 nums 배열에 저장하는 방법이 있더라고요.
다음에 이 방법을 사용해볼 생각입니다.

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