26. Remove Duplicates from Sorted Array

26. Remove Duplicates from Sorted Array

정수 어레이 번호가 감소하지 않는 순서로 정렬된 경우 중복된 요소를 제거하여 각 고유 요소가 한 번만 나타나도록 합니다. 요소의 상대적 순서는 동일하게 유지되어야 합니다. 다른 어레이에 추가 공간을 할당하지 마십시오. 추가 메모리 O(1)를 사용하여 입력 배열이 제자리에서 수정이 되도록 진행해야함.

 

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.

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];
}

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:

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

 

Solution 02.  나의 코드

1. 중복을 제거하는 set()함수 적용 후, sorted()함수로 정리

* 참고로 이 방법은 in-place 요구조건과 맞지 않는다고 함. sorted()와 set()함수는 새로운 배열을 만든다고 함.

원래있던 배열에 넣어주면 num이 아닌 nums[:]에 넣어주는 것이 핵심 !! 이것 알아내느냐고 시간 오지게 걸림

[Runtime:  88ms, Memory Usage:  15.7MB] (Runtime beats 51.57% of python 3 submission)

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

2. previous value를 저장 후 비교하여 pop()하기

[Runtime:  84ms, Memory Usage:  15.7MB] (Runtime beats 66.74% of python 3 submission)

[Memory usage beats 90.44% of python3 submission]

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        prev = -1000
        for i in range(len(nums)-1,-1,-1):
            if prev == nums[i]:
                nums.pop(i)
            prev = nums[i]
        
        return len(nums)