leetcode 80) Remove Duplicates from Sorted Array II

LV. Medium 🧐

https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/

Remove Duplicates from Sorted Array II - LeetCode

 

문제

Given an integer array nums sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. 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 = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,0,1,1,1,1,2,3,3]
Output: 7, nums = [0,0,1,1,2,3,3,_,_]
Explanation: Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

 

Constraints:

• 1 <= nums.length <= 3 * 10^4
• -10^4 <= nums[i] <= 10^4
• nums is sorted in non-decreasing order.

 

문제 해결법

엄청 간단한 문제!

이전 값을 저장하고 다음 값이 이전 값과 같으면 처음 한 번은 flag를 false로 바꾸고 nums에 저장하고 넘어간다.

(flag를 두어 nums에 중복된 값이 2개까지만 저장할 수 있게 한다.)

 

그다음 값도 이전 값과 같으면 flag가 false이니 저장할 수 없다는 것을 알고 nums에 저장하지 않고 넘긴다.

 

해결 코드

class Solution {
public:
	int removeDuplicates(vector<int>& nums) {
		int ans = 1;
		bool flag = true;
		int before = nums[0];

		for (int i = 1; i < nums.size(); ++i) {
			if (before == nums[i] && !flag)
				continue;
			else if (before == nums[i] && flag)
				flag = false;
			else {
				flag = true;
				before = nums[i];
			}
			nums[ans] = nums[i];
			ans++;
		}
		return ans;
	}
};