Remove Duplicates from Sorted Array II

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本文介绍了一种算法解决方案，用于将一个已排序数组中的重复元素限制为最多出现两次。通过具体示例展示了如何从数组[1,1,1,2,2,3]中移除多余的重复项，使得最终数组变为[1,1,2,2,3]，并返回新数组的有效长度。

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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].

Better Approach
Key to solve:
start from index: prev=1; curr=2 since allow twice duplication;
There are comparison cases:
1:Only twice duplication allow: compare with two prior elements:
2. Not accepted case.

Check coding in detail below

public class Solution {
    public int removeDuplicates(int[] A) {
        int len=A.length;
        if(len<3) return len;
        int prev=1;
        int cur=2;
        while(cur<len){
            //accepted case
            if(A[cur]==A[prev] && A[cur]==A[prev-1]){
                cur++;
            }else{  //unaccepted case
                prev++;
                A[prev]=A[cur];
                cur++;
            }
        }
        return prev+1;
    }
}