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].

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Analysis:

We need to do two binary search. The first finds out the left boundary and the second finds out the right boundary.

if mid==target, we then check whether mid-1==target, if yes, we then continue to search [start,mid-1] until we find the left boundary.

Solution:

public class Solution {
    public int[] searchRange(int[] A, int target) {
        int[] res = new int[]{-1,-1};
        if (A.length==0) return res;

        int start = 0, end = A.length-1;       
      
        //Find left range.

        while (start<=end){
            int mid = (start+end)/2;
            if (A[mid]==target){
                //find right range also.
                if (mid+1==A.length || A[mid+1]!=target) res[1] = mid;
               
                //Check left
                if (mid-1==-1 || A[mid-1]!=target){
                    res[0]=mid;
                    break;
                } else {
                    end = mid-1;
                    continue;
                }
            } else if (A[mid]>target) end = mid-1;
            else start = mid+1;
        }

        if (start>end) return res;
        if (res[0]!=-1 && res[1]!=-1) return res;

        //Find right
        start = 0;
        end = A.length-1;
        while (start<=end){
            int mid = (start+end)/2;
            if (A[mid]==target){
                //Check right
                if (mid+1==A.length || A[mid+1]!=target){
                    res[1]=mid;
                    break;
                } else {
                    start = mid+1;
                    continue;
                }
            } else if (A[mid]>target) end = mid-1;
            else start = mid+1;
        }

        return res;        
    }
}