Leetcode126: 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 n = nums.size();        int l = 0;        int r = n-1;        int res = -1;        vector<int> ret;        while(l<=r)        {            int mid = (l+r)/2;            if(nums[mid] == target)            {                res = mid;                break;            }            else if(nums[mid] < target)                l = mid+1;            else                r = mid-1;        }        if(res == -1)        {            ret.push_back(-1);            ret.push_back(-1);              return ret;        }        int ls=res;        int rs=res;        while(ls>=0 && nums[ls] == nums[res])         {            ls--;        }        ls++;        while(rs<=n-1 && nums[rs] == nums[res])        {            rs++;        }        rs--;        ret.push_back(ls);        ret.push_back(rs);        return ret;    }};