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.

已排序数组，加一个变量记录下元素出现次数即可，如果是没有排序的数组，则需要引入一个hashmap来记录出现的次数。

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C++

class Solution {public:    int removeDuplicates(vector<int>& nums) {        if(nums.size() <= 2)            return nums.size();        int index = 2;        for(int i = 2; i < nums.size(); i++)        {            if(nums[i] != nums[index - 2])                nums[index++] = nums[i];        }        return index;    }};

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Java

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

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Python

class Solution(object):    def removeDuplicates(self, nums):        """        :type nums: List[int]        :rtype: int        """        if len(nums) <= 2:            return len(nums)        index = 2        for i in range(2,len(nums)):            if nums[i] != nums[index -2]:                nums[index] = nums[i]                index += 1        return index