Search for a Range

水题一道：
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].

AC的C++代码如下：

    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> ret;        int left = -1, right = -1;        for(int i=0; i<nums.size(); i++){            if(nums[i] == target && left < 0){                left = i;                right = left;            }else if(nums[i] == target){                right = i;            }        }        ret.push_back(left);        ret.push_back(right);        return ret;    }