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].

Analyse: binary search. 

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> range = {-1, -1};
        int left = 0, right = nums.size() - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] > target) right = mid - 1;
            else if (nums[mid] < target) left = mid + 1;
            else {
                int start = mid, end = mid;
                while (start >= 0 && nums[start] == nums[mid]) start--;
                while (end < nums.size() && nums[end] == nums[mid]) end++;
                range[0] = start + 1;
                range[1] = end - 1;
                break;
            }
        }
        return range;
    }
};