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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int min,max,mid;
        int i, j;
        min = 0;
        max = n-1;
        vector<int> result ;
        if(A[0] > target|| A[n-1] <target)
        {
            result.push_back(-1);
            result.push_back(-1);
            return result ;
        }
        while(min <= max)
        {
            mid = min + (max - min)/2;
            if(A[mid] == target)
               break;
             else if(A[mid] < target)
                 min = mid + 1;
               else
                 max = mid - 1;
        }
        
      if( A[mid] == target ) {
            i=j = mid;
            while(i>=0 && A[i] == target)i-- ;
            while(j<= n-1 && A[j] == target) j++ ;
             i++;j--;
        result.push_back(i);
        result.push_back(j) ;
      }else{
       
           result.push_back(-1);
            result.push_back(-1);  
      }
        return result ;
    }
};