26. Remove Duplicates from Sorted Array

개굴·2024년 5월 13일

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.

계획

My plan

for문으로 훑으면서 앞뒤를 비교
기존 list에서 지워버리면 안 되므로 포인터를 두 개 사용

Solution code

python의 set함수 이용 -> set 함수는 중복을 제거한다. 대신 순서가 없음

코드

My plan

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

Solution code

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:        
     
     	nums[:] = sorted(set(nums))
        return len(nums)

Review - 20240702

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        
        num_set = set()

        write = 0

        for i in range(len(nums)):
            if nums[i] not in num_set:
                num_set.add(nums[i])
                nums[write] = nums[i]
                write+=1
        
        return write