/*

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).

*/

public class LeetCode26 {

static public int removeDuplicates(int[] nums) {

int ptr=0,k=0;

for(int i=0;i<nums.length;i++){

ptr = nums[i];

while(i<nums.length && ptr==nums[i] ){

i++;

}

nums[k]=ptr; k++; i--;

}

return k;

}

public static void main(String[] args) {

System.out.println(

);

}

}