**source:**https://leetcode.com/problems/remove-duplicates-from-sorted-array/**C/C++**

**Solution to LeetCode**problem

**26**.

**Remove Duplicates from Sorted Array**.

## Problem

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 lengthint 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**.

## Examples

**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

`1 <= nums.length <= 3 * 10`

^{4}`-100 <= nums[i] <= 100`

`nums`

is sorted in**non-decreasing**order.

## Solution

- We will handle two indexes.
`i`

to iterate through the array of numbers.`p`

to keep tracking of the last non-repeated number.

- Both indexes start at the same position and they move together.
- Once the
`nums[i] == nums[i-1]`

we continue moving the`i`

index until they are different, but`p`

remains in that position. - Once the numbers are different, we will “move” them to the index where
`p`

is, and we move`p`

one place. At this point, we will move all the non repeated numbers to the position of`p`

. - Finally, we return the index
`p`

.

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class Solution {
public:
int removeDuplicates(vector<int>& nums) {
int p=1;
for(int i=1; i<nums.size(); i++) {
if (nums[i] == nums[i-1])
continue;
if (p != i)
nums[p] = nums[i];
p++;
}
return p;
}
};