[Leetcode 34, Medium] Search for a Range

Problem:

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].

Analysis:

Solutions:

C++:

    vector<int> searchRange(vector<int>& nums, int target) {
        if(nums.empty() || target < nums[0] || target > *(nums.end() - 1))
            return vector<int>(2, -1);
        
        vector<int> range_ends;
        int start = 0;
        int end = nums.size() - 1;
        while(start <= end) {
            int mid = (start + end) / 2;
            
            if(nums[mid] == target) {
                int left = mid - 1;
                for(; left >= 0; --left) {
                    if(nums[left] != nums[mid])
                        break;
                }
                range_ends.push_back(left + 1);
                
                int right = mid + 1;
                for(; right < nums.size(); ++right) {
                    if(nums[right] != nums[mid])
                        break;
                }
                range_ends.push_back(right - 1);
                
                break;
            } else if(nums[mid] < target)
                for(start = mid + 1; nums[start] == nums[mid]; ++start) {}
            else
                for(end = mid - 1; nums[end] == nums[mid]; --end) {}
        }
        
        return range_ends.empty() ? vector<int>(2, -1) : range_ends;
    }

Java:

Python: