[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].

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