Search for a Range [LeetCode]

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].

Summary: Just a practice of binary search.

    vector<int> searchRange(int A[], int n, int target) {
        vector<int> range(2,-1);
        int start = 0;
        int end = n - 1;
        while(start <= end) {
            int median = start + (end - start + 1) / 2;
            if(A[median] > target) {
                end = median - 1;
            }else if (A[median] < target) {
                start = median + 1;
            }else { //equals
                //go right
                int i = median + 1;
                for(; i <= end; i ++) {
                    if(A[i] != target){
                        range[1] = i - 1;
                        break;
                    }
                }
                if(range[1] == -1)
                    range[1] = i - 1;
                //go left
                i = median - 1;
                for(; i >= start; i --) {
                    if(A[i] != target){
                        range[0] = i + 1;
                        break;
                    }
                }
                if(range[0] == -1)
                    range[0] = i + 1;
                
                break;
            }
        }
        return range;
    }