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.
Solution：Two pointers
思路：nums[new_i - 2] != nums[i] 则可加
Time Complexity: O(N) Space Complexity: O(1)
Solution Code：
class Solution {
public int removeDuplicates(int[] nums) {
int new_i = 0;
for(int i = 0; i < nums.length; i++) {
if(new_i < 2 || nums[new_i - 2] != nums[i]) {
nums[new_i++]= nums[i];
}
}
return new_i;
}
}
