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 low  = 0;
        int high = n-1;
        int mid  = (low+high)/2;
        
        int beginPos = -1;
        int endPos   = -1;
        
        vector<int> retVector;

        // first, we want to find the left,
        while (mid < high)
        {
            // here is the main proces
            if (A[mid] >= target)
            {
                high = mid;
            }
            else
            {
                low = mid+1;
            }
            mid = (low+high)/2;
        }
        
        if (A[mid] != target)
        {
            retVector.push_back(beginPos);
            retVector.push_back(endPos);           
            return retVector;
        }
        beginPos = mid;
        
        low  = 0;
        high = n-1;
        mid  = (low+high)/2;
        
        while (mid > low)
        { 
            if (A[mid] <= target)
            {
                low = mid;
            }
            else
            {
                high = mid-1;
            }
            mid = (low+high)/2;
        }
        
        if (A[high] == target)
        {
            endPos = high;
        }
        else
        {
            endPos = mid;
        }
        
        retVector.push_back(beginPos);
        retVector.push_back(endPos);
        
        return retVector;
    }
};