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) {        int l = 0, r = n-1;        vector<int> result;        result.push_back(-1);        result.push_back(-1);        while(l <= r)        {            int mid = (l+r)/2;            if(A[mid] == target)            {                int iter = mid;                while(iter >= 0 && A[iter] == target)                {                    iter--;                }                result[0] = iter+1;                iter = mid;                while(iter < n && A[iter] == target)                {                    iter++;                }                result[1] = iter-1;                return result;                            }            else if (A[mid] < target)            {                l = mid+1;            }            else            {                r = mid-1;            }        }        return result;    }};