LeetCode Solutions : Remove Duplicates from Sorted Array I & II

Remove Duplicates from Sorted Array 

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 A = [1,1,2],

Your function should return length = 2, and A is now [1,2].

public class Solution {
    public int removeDuplicates(int[] A) {
        int len=A.length;
        if(len<2) return len;
        int j=0;
        for(int i=1;i<len;){
            if(A[i]==A[j])
                i++;
            else
                A[++j]=A[i++];
        }
        return j+1;
    }
}

Remove Duplicates from Sorted Array II 

Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

For example,
Given sorted array A = [1,1,1,2,2,3],

Your function should return length = 5, and A is now [1,1,2,2,3].

1. The same solution imitating the above one

public class Solution {
	public int removeDuplicates(int[] A) {
		int len=A.length;
		if (len<3)
			return len; 
		int pre = A[0];
		boolean flag = false;
		int count = 0;
		// index for updating
		int j = 1; 
		for (int i = 1; i < len; i++) {
			int cur = A[i]; 
			if (cur == pre) {
				if (!flag) {
					flag = true;
					A[j++] = cur;
					continue;
				} else {
					count++;
				}
			} else {
				pre = cur;
				A[j++] = cur;
				flag = false;
			}
		}
 
		return A.length - count;
	}
}

2. Better solution

public class Solution {
    public int removeDuplicates(int[] A) {
        int len=A.length;
		if(len<3)
			return len;
		int pre=1;
		int cur=2;
		while(cur<len){
			if(A[cur]==A[pre]&&A[cur]==A[pre-1]){ // There are three consecutive same values
				cur++;
			}else{
				pre++;
				A[pre]=A[cur];
				cur++;
			}
		}
		return pre+1;
    }
}