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.

题意：给一个数组，有序的，去掉超过两个的即可。

java代码：

class Solution {
    public int removeDuplicates(int[] nums) {

                int count = 0;
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {

            if (map.get(nums[i]) == null) {
                count++;
                map.put(nums[i], 1);
            } else if (map.get(nums[i]) == 1) {
                count++;
                map.put(nums[i], 2);
            }
        }

        List<Integer> list = new ArrayList<Integer>();
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            if (entry.getValue() == 1) {
                list.add(entry.getKey());
            } else {
                list.add(entry.getKey());
                list.add(entry.getKey());
            }

        }

        Collections.sort(list);

        for (int i = 0; i < list.size(); i++) {
            nums[i] = list.get(i);
        }

        return count;
    }

}