# 26.Remove Duplicates From Sorted Array

**<font color=red>难度Easy</font>**

## 刷题内容

> 原题连接

* https://leetcode.com/problems/remove-duplicates-from-sorted-array/
> 内容描述

```
Given a sorted array nums, 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 by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,2],

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:

Given nums = [0,0,1,1,1,2,2,3,3,4],

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.
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 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]);
}
```
> 思路
******- 时间复杂度: O(N)******- 空间复杂度: O(1)******

由于数组已经排序好的，我们直接遍历数组，用一个指针 i 指向开始处，另一个 j 指向第二个，若 j 指向的数不等于 i 指向的数，则```++i,++j```。否则只```++j```

```cpp
class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int j = 0;
        if(!nums.size())
            return 0;
        for(int i = 1;i < nums.size();)
        {
            while(nums[j] == nums[i] && i < nums.size())
               ++i;
            if(i == nums.size())
                break;
            swap(nums[++j],nums[i++]);
        }
        return j + 1;
        }
};
```