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。

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;
}
};

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;
}
}

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