Follow up for "Remove Duplicates":

What if duplicates are allowed at most twice?

For example,

Given sorted array 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. It doesn't matter what you leave beyond the new length.

 

典型的双指针问题，维护两个指针不断更新就可以了，代码如下：

1 class Solution { 2 public: 3     int removeDuplicates(vector
     
      & nums) { 4         int sz = nums.size(); 5         if(!sz) return 0; 6         int start = 0;//第二个指针 7         int i = 0;//第一个指针 8         int count = 0; 9         int key = nums[0];10         for(; i < sz; ++i){11             if(nums[i] == key)12                 count++;13             else{14                 for(int k = 0; k < min(2, count); ++k)15                     nums[start++] = key;//更改数组元素，指针前移16                 key = nums[i];17                 count = 1;18             }19         }20         for(int k = 0; k < min(2, count); ++k)21             nums[start++] = key;22         return start;23     }24 };