[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 order of O(log n).
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].

The variation of traditional binary search. In order to be consisted with the version of binary search I'm most comfortable with, I used one kinds of the implementation that is most similar with the original one to find the first occurrence and the last occurrence . There are many other ways of implementing the functionality and you need only to recite one of them.

public class Solution {
    public int[] searchRange(int[] A, int target) {
        int[] res = new int[2];
        res[0] = left(A, target);
        res[1] = right(A, target);
        return res;
    }
    
    private int left(int[] A, int target) {
        int l = 0, r = A.length-1, result = -1;
        while(l <= r) {
            int mid = l + (r - l) / 2;
            if(A[mid] == target) {
                result = mid;
                r = mid - 1;
            }
            else if(A[mid] > target)
                r = mid - 1;
            else
                l = mid + 1;
        }
        return result;
    }
    
    private int right(int[] A, int target) {
        int l = 0, r = A.length-1, result = -1;
        while(l <= r) {
            int mid = l + (r - l) / 2;
            if(A[mid] == target) {
                result = mid;
                l = mid + 1;
            }
            else if(A[mid] > target)
                r = mid - 1;
            else
                l = mid + 1;
        }
        return result;
    }
}