LeetCode 80. Remove Duplicates from Sorted Array II

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.

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有序数组去重，每个重复元素最多保留两个。
和LeetCode 26.Remove Duplicates from Sorted Array
相识，只不过这里每个重复的元素可以保留两个。

class Solution {
    public int removeDuplicates(int[] nums) {
        int i,j;
         for(i=2,j=2 ; i<nums.length;i++)
             if(nums[j-2]!=nums[i]) 
                 nums[j++]=nums[i];
         return j;
    }
}

这类问题有个总结的解法 如果是可以保留重复k个元素

class Solution {
    public int removeDuplicates(int[] nums，int k) {
        if(nums.size()<k) return nums.size(); // if array size is less than k then return the same
        int i,j;
         for(i=k,j=k ; i<nums.length;i++)
             if(nums[j-k]!=nums[i]) nums[j++]=nums[i];
         return j;
    }