难度：Medium
解题思路：本题是remove_duplicates_from_sorted_array的一个升级版。同样使用两个迭代器，一个指向已经处理好的数组的最后一位，另一个进行遍历。需要加入一个计数器，存储当前元素已经处理过的个数。
代码如下：
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class Solution {
public:
int removeDuplicates(vector<int>& nums) {
auto solved_it = nums.begin();
if(nums.size() <= 2) return nums.size();
auto unsolved_it = solved_it+1;
int count = 1;
while(unsolved_it != nums.end())
{
if(*unsolved_it != *solved_it && (count == 2 || count == 1))
{
swap(*unsolved_it, *(++solved_it));
count=1;
unsolved_it++;
}
else if(*unsolved_it == *solved_it && count == 2)
{
unsolved_it++;
}
else if(*unsolved_it == *solved_it && count == 1)
{
swap(*unsolved_it, *(++solved_it));
count=2;
unsolved_it++;
}
}
return solved_it-nums.begin()+1;
}
};
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运行结果：19ms,超过15.84%
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