問題：
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.

這個問題相對而言比較簡單，在紙上畫1下處理進程便可。

代碼示例：

public class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums == null || nums.length < 1) {
            return 0;
        }

        int begin = 0;
        int end = 1;

        //記錄同1個字符重復的次數(shù)
        int count = 1;

        while (end < nums.length) {
            //end與begin對應數(shù)1致時
            if (nums[end] == nums[begin]) {
                //更新count
                ++count;

                //count <= 2時，將end移動到begin后1個位置，同時增加begin
                //否則，數(shù)量過量，不移動begin，直到找到下1個不1樣的數(shù)
                if (count <= 2) {
                    nums[begin+1] = nums[end];
                    ++begin;
                }
            } else {
                //找到不1樣的數(shù)，將end移動到begin后1個位置，同時增加begin
                count = 1;
                nums[begin+1] = nums[end];
                ++begin;
            }
            ++end;
        }

        return begin+1;
    }
}