[LeetCode] Arrays101 - Remove Duplicates from Sorted Array

Remove Duplicates from Sorted Array

Given a sorted array nums, remove the duplicates in-place such that each element appears only once and returns the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

 

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means a modification to the input array will be known to the caller as well.

Internally you can think of this:

 

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

 

Example 1:

Input: nums = [1,1,2]
Output: 2, nums = [1,2]
Explanation: Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the returned length.

 

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 length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively. It doesn't matter what values are set beyond the returned length.

 

Constraints:

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

---

 

Remove Elements 문제로 고생을 많이 했더니 비슷한 문제는 비교적 금방 풀었다!

 

일단 배열이 정렬이 된 상태로 들어오기 때문에

pointer를 2개로 두고 값이 변경되는 순간을 인지해서 값을 덮어쓰기 하면 된다.

 

/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function(nums) {
    let i = 0;
    
    for (let j = 1; j < nums.length; j++) {
        if (nums[j] !== nums[i]) {
            nums[++i] = nums[j];
        }
    }
    
    return i + 1;
};