80. Remove Duplicates from Sorted Array II（python+cpp）

题目：

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice 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 = [1,1,1,2,2,3],
Your function should return length = 5, with the first five elements of nums being 
1, 1, 2, 2 and 3 respectively.
It doesn't matter what you leave beyond the returned length. 

Example2:

Given nums = [0,0,1,1,1,1,2,3,3],
Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 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 thefirst len elements. 
for (int i = 0; i < len; i++) {
    print(nums[i]); 
}

解释：
本来以为是类似于82. Remove Duplicates from Sorted List II（python+cpp）的要求，结果发现不是的。
题目要求：给定一个排好序的数组，in-place删除一些重复元素使得最终返回的数组多种每个数组至多出现两次。返回最终的数组的长度。
其实这种题目也是需要用两个指针的，先写一个不那么明显的用指针的方法，速度比较慢。
pytho代码：

class Solution:
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        i=0
        for n in nums:
            if i<2 or n>nums[i-2]:
                nums[i]=n
                i+=1
        return i

c++代码：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        //j指向当前需要赋值的位置
        int j=0;
        for(int i=0;i<nums.size();i++)
        {
            if (i<2 || nums[j-2]<nums[i])
            {
                nums[j]=nums[i];
                j+=1;
            }
        }
        return j;
    }
};

用两个指针的解法：
python代码：

class Solution:
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n=len(nums)
        if n<=2:
            return n
        count,i,j=1,1,1
        while i< n:
            if nums[i]==nums[i-1]:
                if count<2:
                    count+=1
                    nums[j]=nums[i]
                    j+=1
                i+=1
            else:
                count=1
                nums[j]=nums[i]
                i+=1
                j+=1
        return j     

c++代码：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int n=nums.size();
        if(n<=2)
            return n;
        int count=1,i=1,j=1;
        while(i<n)
        {
            if(nums[i]==nums[i-1])
            {
                if (count<2)
                {
                    count++;
                    nums[j]=nums[i];
                    j++;
                }
            }
            else
            {
                count=1;
                nums[j]=nums[i];
                j++; 
            }
            i++;
        }
        return j;
    }
};

总结：
注意变量的初始化和一些特殊情况的判断。