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].

class Solution {
public:
    
    vector<int> searchRange(int A[], int n, int target) {
        vector<int>fail;
        fail.push_back(-1);
        fail.push_back(-1);
        vector<int>ret;
         if(n<=0){
            return fail; 
         }
         if(n==1){
             if(A[0]==target){
                 ret.push_back(0);
                 ret.push_back(0);
                 return ret;
             }
             else return fail;
         }
         int index1=-1,index2=n;
        //find the last element less than target
        int start,end;
        start=0;end=n-1;
        while(start<=end){
            int mid=(start+end)/2;
            if(A[mid]>=target){
                end=mid-1;
            }
            else{
                if(mid==n-1){
                    index1=mid;
                    break;
                }
                if(A[mid+1]>=target){
                    index1=mid;
                    break;
                }
                else{
                    start=mid+1;
                }
            }
        }
        //find the first element great than target
        start=0;end=n-1;
        while(start<=end){
            int mid=(start+end)/2;
            if(A[mid]<=target){
                start=mid+1;
            }
            else{
                if(mid==0){
                    index2=0;
                    break;
                }
                else{
                    if(A[mid-1]<=target){
                        index2=mid;
                        break;
                    }
                    else{
                        end=mid-1;
                    }
                }
                
            }
        }
        if(index1==index2-1)return fail;
        else{
            ret.push_back(index1+1);
            ret.push_back(index2-1);
            return ret;
        }
    }
};