Problem
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
numssuch that the firstkelements ofnumscontain the unique elements in the order they were present innumsinitially. The remaining elements ofnumsare not important as well as the size ofnums. - 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.
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] <= 100numsis sorted in non-decreasing order.
Approach
EN
It was very similar to the last problem(Remove Element). But this problem doesn't have an argument value. So I created one which is nums[i+1]. The rest will be same. If val and nums[i] is the same, nums[i] needs to be removed from nums array. Last, return the length of the nums array.
KR
바로 전 문제 (Remove Element)와 굉장히 유사한 문제였다. 다만 이번엔 아규먼트로 들어오는 val이 없고, 중복된 원소없이 유니크한 값들로만 이루어져 있어야 한다. 그렇다면 val을 내가 만들어주면 되겠구나! val은 nums[i+1]로 잡고 계속 비교하며 검사했다. 만약 val과 nums[i] 값이 같다면 똑같이 nums 배열에서 제거해주고 반복문을 돌고 있는 인덱스도 줄여준다. 마지막으로, 줄어든 nums 배열의 길이를 리턴한다.
code
/**
* @param {number[]} nums
* @return {number}
*/
var removeDuplicates = function(nums) {
for(let i=0; i<nums.length; i++) {
let val = nums[i+1];
if(nums[i]==val) {
nums.splice(i, 1);
i--;
}
}
return nums.length();
};