LeetCode 솔루션

[Easy - wk2 - Q2] 26. Remove Duplicates from Sorted Array

Chloe 쪽지보내기 메일보내기 자기소개 아이디로 검색 전체게시물 작성
작성일 2022.05.08 20:09

7,525 조회
5 댓글
0 추천
0 비추천
목록 게시글 쓰기

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];
}

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.

#leetcode, #문제풀이, #easy, #facebook, #amazon, #adobe, #microsoft, #linkedin, #bloomberg, #google, #apple, #vmware, #walmart global tech, #array, #two pointers

익명
작성일 2022.05.10 10:36

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        cur = 0
        for i in range(1, len(nums)):
            if nums[cur] != nums[i]:
                cur+=1
                nums[cur] = nums[i]
          
        return cur+1

익명
작성일 2022.05.11 20:04

        count = 0
        for i in range(len(nums)):
            if nums[i] != nums[count]:
                count+=1
                nums[count] = nums[i]
        return count+1

익명
작성일 2022.05.11 19:29

pytnon

Runtime: 102 ms, faster than 70.67% of Python3 online submissions for Remove Duplicates from Sorted Array.
Memory Usage: 15.5 MB, less than 62.17% of Python3 online submissions for Remove Duplicates from Sorted Array.

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        if len(nums) == 0 : return 0
        if len(nums) == 1 : return 1
        
        x = 1
        
        for i in range(1, len(nums)):
            if nums[i] != nums[i-1]:
                nums[x] = nums[i]
                x += 1
        
        nums[:]=nums[:x]
        
        return x

익명
작성일 2022.05.18 07:22

var removeDuplicates = function(nums) {
 let output = nums.length;
  
  for (let i = 0; i < nums.length; i++) {
      if (nums[i] === nums[i+1]) {
          nums.splice(i, 1);
          output-=1;
          nums.length-1;
          i--;
      }
  }

  let nullCounter = output;
  while (nullCounter > 0) {
    nums.push(null);
    nullCounter-=1;
  }
  return output;
};

익명
작성일 2022.05.18 22:07

C++

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int n = nums.size();
        int index = 0;

        if (n == 0)
            return 0;
        for (int i = 1; i < n; i++)
            if (nums[i] != nums[index])
                nums[++index] = nums[i];
        return index + 1;
    }
};

Python

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        counter = 0
        for i in range(1, len(nums)):
            if nums[counter] != nums[i]:
                nums[counter + 1], nums[i] = nums[i], nums[counter + 1]
                counter += 1
        return counter + 1 if nums else 0