(Leet Code c++)Remove Duplicates from Sorted Array

26. Remove Duplicates from Sorted Array

 

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:

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

 

주어진 nums 벡터만을 이용해 중복된 숫자를 지운 후, nums의 크기를 리턴해야 한다.

in-place이기 때문에 nums만을 사용.

for문을 통해 i번째 수와 i + 1의 수가 같은지 확인하여 중복이 있으면 i번째를 지우도록 한다.

단, i + 1은 nums.size()보다 작아야 한다.

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        for(int i = 0; i < nums.size(); i++){
            if(i + 1 < nums.size()){
                if(nums[i] == nums[i + 1]){
                    nums.erase(nums.begin() + i);
                    i--;
                }
            }
        }
        
        return nums.size();
    }
};