LeetCode – Remove Duplicates from Sorted Array II (Java)

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].

Naive Approach

Given the method signature "public int removeDuplicates(int[] A)", it seems that we should write a method that returns a integer and that's it. After typing the following solution:

public class Solution {    public int removeDuplicates(int[] A) {        if(A == null || A.length == 0)            return 0;         int pre = A[0];        boolean flag = false;        int count = 0;         for(int i=1; i<A.length; i++){            int curr = A[i];             if(curr == pre){                if(!flag){                flag = true;                    continue;                }else{                    count++;                }            }else{                pre = curr;                flag = false;            }        }         return A.length - count;    }}

Online Judge returns:

Submission Result: Wrong AnswerInput:[1,1,1,2]Output:[1,1,1]Expected:[1,1,2]

So this problem also requires in-place array manipulation.

Correct Solution

We can not change the given array's size, so we only change the first k elements of the array which has duplicates removed.

public class Solution {public int removeDuplicates(int[] A) {if (A == null || A.length == 0)return 0; int pre = A[0];boolean flag = false;int count = 0; // index for updatingint o = 1; for (int i = 1; i < A.length; i++) {int curr = A[i]; if (curr == pre) {if (!flag) {flag = true;A[o++] = curr; continue;} else {count++;}} else {pre = curr;A[o++] = curr;flag = false;}} return A.length - count;}}

Better Solution

public class Solution {public int removeDuplicates(int[] A) {if (A.length <= 2)return A.length; int prev = 1; // point to previousint curr = 2; // point to current while (curr < A.length) {if (A[curr] == A[prev] && A[curr] == A[prev - 1]) {curr++;} else {prev++;A[prev] = A[curr];curr++;}} return prev + 1;}}