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].

Solution:

Code:

<span style="font-size:14px;">class Solution {public:    void binarySearch(int A[], int begin, int end, const int &target, vector<int> &result) {        if (begin <= end) {            int mid = (begin+end)/2;            if (A[mid] == target) {                result[0] = result[0]==-1?mid:min(result[0], mid);                result[1] = max(result[1], mid);                binarySearch(A, mid+1, end, target, result);                binarySearch(A, begin, mid-1, target, result);            } else if (A[mid] < target) binarySearch(A, mid+1, end, target, result);            else binarySearch(A, begin, mid-1, target, result);        }    }        vector<int> searchRange(int A[], int n, int target) {        vector<int> result(2, -1);        binarySearch(A, 0, n-1, target, result);        return result;    }};</span>