_26.java

package com.fishercoder.solutions;

/** Given a sorted array, remove the duplicates
 * in place such that each element appear only once and return the new length.
 * Do not allocate extra space for another array, you must do this in place with constant memory.

 For example,
 Given input array A = [1,1,2],

 Your function should return length = 2, and A is now [1,2].*/

public class _26 {

    public static int removeDuplicates_editorial_solution(int[] nums) {
        int i = 0;
        for (int j = 1; j < nums.length; j++) {
            if (nums[i] != nums[j]) {
                i++;
                nums[i] = nums[j];
            }
        }
        return i + 1;
    }


    public static void main(String... strings) {
        int[] nums = new int[]{1, 1, 2};
//        int[] nums = new int[]{1,1,2,2,3};
//        int[] nums = new int[]{1,1};
        System.out.println(removeDuplicates_editorial_solution(nums));
    }

    /**
     * Same idea as the editorial solution, mine just got more verbose.
     */
    public static int removeDuplicates_my_original(int[] nums) {
        int i = 0;
        for (int j = i + 1; i < nums.length && j < nums.length; ) {
            while (j < nums.length && nums[i] == nums[j]) {
                j++;
            }
            if (j == nums.length) {
                j--;
            }
            int temp = nums[j];
            nums[j] = nums[i + 1];
            nums[i + 1] = temp;
            if (nums[i] != nums[i + 1]) {
                i++;
            }
            if (j == nums.length) {
                break;
            }
            j++;
        }
        return i + 1;
    }

}