[LeetCode] 34. Search for a Range

Given an array of integers sorted in ascending order, 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].

//better one from discuss, qp use l, r
class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> ret(2, -1);
        int n = nums.size();
        int l = 0;
        int r = n - 1;
        int mid;
        
        while (l < r){
            mid = (l+r) /2;
            if (target > nums[mid]) l = mid + 1;
            else if (target <= nums[mid]) r = mid;
            //else r = mid; // push to left
        }
        
        if (nums[l] != target) return ret;
        else ret[0] = l;
        
        r = n - 1; // no need to reinit l;
        while (l < r){
            /*  !!!! make the r biasd to right */
            mid = (l+r+1) / 2;
            if (target < nums[mid]) r = mid - 1;
            else if (target >= nums[mid]) l = mid;
            //else l = mid; // move to right
        }
        
        ret[1] = l;
        
        return ret;
    }
};