Leetcode[26, 80] Remove Duplicates from Sorted Array

###Task1 Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example, Given input array nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn’t matter what you leave beyond the new length.

###Python


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

###Java


public class Solution {
    public int removeDuplicates(int[] nums) {
        //edge cases
    	if (nums.length < 2) return nums.length;
    	
        int index = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] != nums[i - 1]) {
                nums[index] = nums[i];
                index++;
            }
        }
        
        return index;
    }
}

###Task2 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.

###Python


class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if nums == None or len(nums) == 0:
            return 
        cur = nums[0]
        count = 1
        ind = 1
        for i in range(1, len(nums)):
            if nums[i] == cur and count >= 2:
                continue
            elif nums[i] == cur and count == 1:
                count += 1
                nums[ind] = nums[i]
                ind += 1
            else:
                cur = nums[i]
                nums[ind] = nums[i]
                count = 1
                ind += 1
        return ind
            

###Java


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

 		int idx = 0;
 		int count = 0;

 		for (int i = 0; i < nums.length; i++) {
 			if (i > 0 && nums[i] == nums[i - 1]) {
 				count++;
 				if (count > 2) {
 					continue;
 				}
 			} else {
 				count = 1;
 			}

 			nums[idx++] = nums[i];
 		}

 		return idx;
    }
}

###Points