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].

vector<int> searchRange(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int begin = 0;
		int end = n-1;
		int index = -1;
		vector<int> res;
		while(begin < end)
		{
			int mid = (begin+end)/2;
			if(A[mid] == target){index = mid; break;}
			else if(target < A[mid])
				end = mid-1;
			else
				begin = mid+1;
		}
		if(begin==end && A[begin]==target)
			index = begin;
		if(index == -1)
		{
			res.push_back(-1);
			res.push_back(-1);
			return res;
		}else{
			begin = index;
			while(begin >=0 && A[begin]==target)begin--;
			begin++;
			end = index;
			while(end < n && A[end]==target)end++;
			end--;
			res.push_back(begin);
			res.push_back(end);
			return res;
		}
    }