leetCode - Remove Duplicates from Sorted Array

문제

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.

 

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풀이

• 문제 조건을 잘 보아야 한다. 
• 다른 배열을 메모리에 할당 하면 안된다. 
• 주어진 배열에서 가공하여 처리하여야 함

const removeDuplicates = function(nums) {
    let j =0
    for (let i = 0; i < nums.length; i++) {
        nums[j++] = nums[i] 
        while (true) {
            if (nums[i] === nums[i + 1]) {
                i++;
            } else {
                break;
            }
        }
    }
    nums.splice(j, nums.length);
    return nums;
};

 

array.splice(start, deleteCount, [추가할 요소])-새로운 배열을 만드는 것이 아니라 배열의 기존 요소를 교체 또는 삭제

 

반면,

 

array.slice(start, end)-얕은 복사로 새로운 배열을 반환