26. Remove Duplicates from Sorted Array

Method 1

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 nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums.
- 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).

Method 1

【O(n) time | O(1) space】

package Leetcode.TwoPointer;

/**
 * @author zhengxingxing
 * @date 2024/11/14
 */
public class RemoveDuplicatesFromSortedArray {
    public static int removeDuplicates(int[] nums) {
        // If the array is null or has no elements, return 0 as there are no elements to process
        if (nums == null || nums.length == 0) {
            return 0;
        }

        // Initialize a variable k to track the position of the next unique element; start from 1 as the first element is always unique
        int k = 1;

        // Loop through the array starting from the second element
        for (int i = 1; i < nums.length; i++) {
            // If the current element is not equal to the previous one, it is unique
            if (nums[i] != nums[i - 1]) {
                // Place the unique element at the position k, then increment k
                nums[k++] = nums[i];
            }
        }

        // Return the count of unique elements, which is stored in k
        return k;
    }

    public static void main(String[] args) {

        // Test array with sorted elements including duplicates
        int[] nums = {0, 0, 1, 1, 1, 2, 2, 3, 3, 4};

        // Call removeDuplicates to get the count of unique elements and modify the array in place
        int k = removeDuplicates(nums);

        System.out.println("Unique count: " + k);
        System.out.print("Modified array: ");
        for (int i = 0; i < k; i++) {
            System.out.print(nums[i] + " ");
        }
    }
}

Method 1

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