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# Remove Duplicates from Sorted Array
## Remove Duplicates from Sorted Array
原题链接：
### 描述
> Given a sorted array *nums*, remove the duplicates [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) 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**](https://en.wikipedia.org/wiki/In-place_algorithm) with O(1) extra memory.
### 例子
> 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.
* 注意
> 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.
### 解法一
使用index，在原数组上做remove
```cpp
class Solution {
public:
int removeDuplicates(vector& nums) {
if (nums.empty()) return 0;
int index = 0;
for (int i = 1; i < nums.size(); i++) {
if (nums[index] != nums[i]) nums[++index] = nums[i];
}
return index + 1;
}
};
```
或者
```cpp
class Solution {
public:
int removeDuplicates(vector& nums) {
if (nums.empty()) return 0;
int index = 0;
int n=nums.size();
for(int i = 1; i < n; i++){
if(nums[i] == nums[i-1]) index++;
else nums[i-index] = nums[i];
}
return n-index;
}
};
```
**分析：**
T=O(n), S=O(1)
### 解法二
使用STL的distance和unique
**关于distance：**
[**http://cplusplus.com/reference/iterator/distance/?kw=distance**](http://cplusplus.com/reference/iterator/distance/?kw=distance)
Calculates the number of elements between first and last.
**关于unique：**
[**http://cplusplus.com/reference/algorithm/unique/?kw=unique**](http://cplusplus.com/reference/algorithm/unique/?kw=unique)
Removes all but the first element from every consecutive group of equivalent elements in the range `[first,last)`.
```cpp
class Solution {
public:
int removeDuplicates(vector& nums) {
return distance(nums.begin(), unique(nums.begin(), nums.end()));
}
};
```
**分析：**
根据STL中unique是使用双指针实现的：
```cpp
template
ForwardIterator unique (ForwardIterator first, ForwardIterator last)
{
if (first==last) return last;
ForwardIterator result = first;
while (++first != last)
{
if (!(*result == *first)) // or: if (!pred(*result,*first)) for version (2)
*(++result)=*first;
}
return ++result;
}
```
### 解法三
改写unique，使用upper\_bound
```cpp
class Solution {
public:
int removeDuplicates(vector& nums) {
return distance(nums.begin(), removeDuplicates(nums.begin(),nums.end(), nums.begin()));
}
templateOutIt removeDuplicates(InIt first, InIt last, OutIt output) {
while (first != last) {
*output++ = *first;
first = upper_bound(first, last, *first);
}
return output;
}
};
```
**分析：**
## 总结
题目开始，联想到快慢指针的方法。如果对STL熟悉，可以想到unique函数来进行解题，但是题目提示为sorted array，所以unique的解法不一定是最优解。在理解unique的原理和分析题目的基础上，作出变形改写。
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