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].

class Solution {
private:
    int left_bound(std::vector<int> &nums, int target){
        int begin = 0;
        int end = nums.size()-1;
        while(begin<=end)
        {
            int mid = (begin + end)/2;
            if(target == nums[mid])
            {
                if (mid == 0 || nums[mid-1] < target )
                    return mid;
                end = mid - 1;
            }else if(target < nums[mid])
            {
                end = mid - 1;
                
            }else if(target > nums[mid])
            {
                begin = mid + 1;
            }
            
        }
        return -1;
    }
    
    int right_bound(std::vector<int> &nums, int target){
        int begin = 0;
        int end = nums.size()-1;
        while(begin<=end)
        {
            int mid = (begin + end)/2;
            if(target == nums[mid])
            {
                if (mid == nums.size()-1 || nums[mid+1] > target )
                    return mid;
                begin = mid + 1;
            }else if(target < nums[mid])
            {
                end = mid - 1;
                
            }else if(target > nums[mid])
            {
                begin = mid + 1;
            }
            
        }
        return -1;
    }
    
public:
    vector<int> searchRange(vector<int>& nums, int target) {
       vector<int> res(2);
       res[0] = -1;
       res[1] = -1;
       if(nums.size()==0)
           return res;
       res[0] = left_bound(nums, target);
       res[1] = right_bound(nums,  target);
       return res;
    }
};