[leetcode] search for a range

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 searchRange(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
       
        // binary search in the given array
        int start = 0, end = n-1, mid;
        vector result;
        bool is_found = false;
        int c1 = -1, c2 = -1;
       
        while (start <= end)
        {
            mid = (start + end) / 2;
           
            if (target == A[mid])
            {
                is_found = true;
                break;
            }
            else if (target > A[mid])
                start = mid + 1;
            else
                end = mid - 1;
           
        }
       
        if (is_found)
        {
            // move cursors
            c1 = mid, c2 = mid;
            while (A[c1] == target && c1 >= 0)
                c1--;
            while (A[c2] == target && c2 < n)
                c2++;
            c1++;
            c2--;
        }
       
        result.push_back(c1);
        result.push_back(c2);
       
        return result;
    }
};