Leetcode search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the orderO(LogN).

If the target is not found in the array, return[-1, -1].

For example,
Given[5, 7, 7, 8, 8, 10]And target value 8,
Return[3, 4].

Binary Search, although AC, is not well written.

public class Solution {    ArrayList<Integer> arrayList=new ArrayList<>();    int[] res=new int[]{-1,-1};    public int[] searchRange(int[] A, int target) {        if (A.length==0) {            return null;        }        int middle = (A.length-1)/2;        int end=A.length-1;        if (A[middle]>target) {            search(A, 0, middle, target);        }else if (A[middle]<target) {            search(A, middle, end, target);        } else {            search(A, 0, middle, target);            search(A, middle+1, end, target);        }                if (arrayList.isEmpty()) {            return res;        }else {            res[0]=arrayList.get(0);            res[1]=arrayList.get(arrayList.size()-1);            return res;        }    }        private void search(int[] A,int begin,int end,int target) {        if(begin>end){            return;        }        if (begin==end) {            if (A[begin]==target) {                if (!arrayList.contains(begin)) {                    arrayList.add(begin);                }                                        }            return;        }    int middle = (begin+end)/2;    if (A[middle]>target) {        search(A, begin, middle, target);        return;    }    if (A[middle]<target) {        search(A, middle+1, end, target);        return;    }    if (A[middle]==target) {        search(A, begin, middle, target);        search(A, middle+1, end, target);        return;    }    }}

 

Leetcode search for a Range