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]);
}

題目翻譯：

      大概的意思是給一個vector<int> &nums，這個nums是有序的，現在需要把nums中的重複的數字都刪除掉，然後返還這個nums的長度。要求是不能申請額外的儲存空間。

如果輸入為 nums={1,2,3,4,4,5}  返還為n=5，但是要求nums要變成{1,2,3,4,5}

解題思路：

      思路比較簡單，只需要遍歷一遍即可，唯一需要注意的是，如果it=num.erase(it)等函式的使用。

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
	int removeDuplicates(vector<int> &nums)
	{
		if (nums.size() <= 1)
			return nums.size();

		auto it = nums.begin()+1;
		while ( it != nums.end() )
		{
			if (*it == *(it - 1))
				it = nums.erase(it);		
			else
				it++;
		}
		return nums.size();
	}
};


int main()
{
	vector<int> nums = {1,2,3,4,5,5,6,7};

	Solution so;
	int a = so.removeDuplicates(nums);
	cout << a << endl;

	system("pause");
	return 0;

}